Rd Sharma XII Vol 1 2020 for Class 12 Commerce Maths Chapter 10 - Differentiation

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Rd Sharma XII Vol 1 2020 Solutions for Class 12 Commerce Maths Chapter 10 Differentiation are provided here with simple step-by-step explanations. These solutions for Differentiation are extremely popular among Class 12 Commerce students for Maths Differentiation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2020 Book of Class 12 Commerce Maths Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2020 Solutions. All Rd Sharma XII Vol 1 2020 Solutions for class Class 12 Commerce Maths are prepared by experts and are 100% accurate.

Page No 10.103:

Question 1:

Find $\frac{d y}{d x}$ , when

$x = a t^{2} and y = 2 a t$

Answer:

$We have, x = a t^{2} and y = 2 a t \Rightarrow \frac{d x}{d t} = 2 a t and \frac{d y}{d t} = 2 a ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 a}{2 a t} = \frac{1}{t}$

Page No 10.103:

Question 2:

Find $\frac{d y}{d x}$ , when

$x = a (θ + \sin θ) and y = a (1 - \cos θ)$

Answer:

$We have, x = a (θ + \sin θ) and y = a (1 - \cos θ)$

$\Rightarrow \frac{d x}{d θ} = a (1 + \cos θ) and \frac{d y}{d θ} = a \sin θ$

$∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a \sin θ}{a (1 + \cos θ)} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} θ} = \tan \frac{θ}{2}$

Page No 10.103:

Question 3:

Find $\frac{d y}{d x}$ , when

$x = a \cos θ and y = b \sin θ$

Answer:

$We have, x = a \cos θ and y = b \sin θ \Rightarrow \frac{d x}{d θ} = - a \sin θ and \frac{d y}{d θ} = b \cos θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{b \cos θ}{- a \sin θ} = - \frac{b}{a} \cot θ$

Page No 10.103:

Question 4:

Find $\frac{d y}{d x}$ , when

$x = a e^{θ} (\sin θ - \cos θ), y = a e^{θ} (\sin θ + \cos θ)$

Answer:

$We have, x = a e^{θ} (\sin θ - \cos θ) and y = a e^{θ} (\sin θ + \cos θ)$

$\Rightarrow \frac{d x}{d θ} = a [e^{θ} \frac{d}{d θ} (\sin θ - \cos θ) + (\sin θ - \cos θ) \frac{d}{d θ} (e^{θ})] and \frac{d y}{d θ} = a [e^{θ} \frac{d}{d θ} (\sin θ + \cos θ) + (\sin θ + \cos θ) \frac{d}{d θ} (e^{θ})] \Rightarrow \frac{d x}{d θ} = a [e^{θ} (\cos θ + \sin θ) + (\sin θ - \cos θ) e^{θ}] and \frac{d y}{d θ} = a [e^{θ} (\cos θ - \sin θ) + (\sin θ + \cos θ) e^{θ}] \Rightarrow \frac{d x}{d θ} = a [2 e^{θ} \sin θ] and \frac{d y}{d θ} = a [2 e^{θ} \cos θ]$

$∴ \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a (2 e^{θ} \cos θ)}{a (2 e^{θ} \sin θ)} = \cot θ$

Page No 10.103:

Question 5:

Find $\frac{d y}{d x}$ , when

$x = b \sin^{2} θ and y = a \cos^{2} θ$

Answer:

$We have, x = b \sin^{2} θ and y = a \cos^{2} θ ∴ \frac{d x}{d θ} = \frac{d}{d θ} (b \sin^{2} θ) = 2 b \sin θ \cos θ and, \frac{d y}{d θ} = \frac{d}{d θ} (a \cos^{2} θ) = - 2 a \cos θ \sin θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{- 2 a \cos θ \sin θ}{2 b \sin θ \cos θ} = - \frac{a}{b}$

Page No 10.103:

Question 6:

Find $\frac{d y}{d x}$ , when

$x = a (1 - \cos θ) and y = a (θ + \sin θ) a t θ = \frac{π}{2}$

Answer:

$We have, x = a (1 - \cos θ) a n d y = a (θ + \sin θ) ∴ \frac{d x}{d θ} = \frac{d}{d θ} [a (1 - \cos θ)] = a (\sin θ) and \frac{d y}{d θ} = \frac{d}{d θ} [a (θ + \sin θ)] = a (1 + \cos θ) ∴ {[\frac{d y}{d x}]}_{θ = \frac{π}{2}} = {[\frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}]}_{θ = \frac{π}{2}} = {[\frac{a (1 + \cos θ)}{a (\sin θ)}]}_{θ = \frac{π}{2}} = \frac{a (1 + 0)}{a} = 1$

Page No 10.103:

Question 7:

Find $\frac{d y}{d x}$ , when

$x = \frac{e^{t} + e^{- t}}{2} and y = \frac{e^{t} - e^{- t}}{2}$

Answer:

$We have, x = \frac{e^{t} + e^{- t}}{2} and y = \frac{e^{t} - e^{- t}}{2}$

$\Rightarrow \frac{d x}{d t} = \frac{1}{2} [\frac{d}{d t} (e^{t}) + \frac{d}{d t} (e^{- t})] and \frac{d y}{d t} = \frac{1}{2} [\frac{d}{d t} (e^{t}) - \frac{d}{d t} e^{- t}] \Rightarrow \frac{d x}{d t} = \frac{1}{2} [e^{t} + e^{- t} \frac{d}{d t} (- t)] and \frac{d y}{d t} = \frac{1}{2} [e^{t} - e^{- t} \frac{d}{d t} (e^{- t})] \Rightarrow \frac{d x}{d t} = \frac{1}{2} (e^{t} - e^{- t}) = y and \frac{d y}{d t} = \frac{1}{2} (e^{t} + e^{- t}) = x ∴ \frac{d y}{d t} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{x}{y}$

Page No 10.103:

Question 8:

Find $\frac{d y}{d x}$ , when

$x = \frac{3 a t}{1 + t^{2}}, and y = \frac{3 a t^{2}}{1 + t^{2}}$

Answer:

$We have, x = \frac{3 a t}{1 + t^{2}}$
Differentiating with respect to t,
$\frac{d x}{d t} = [\frac{(1 + t^{2}) \frac{d}{d t} (3 a t) - 3 a t \frac{d}{d t} (1 + t^{2})}{{(1 + t^{2})}^{2}}] [using quotient rule] \Rightarrow \frac{d x}{d t} = [\frac{(1 + t^{2}) (3 a) - 3 a t (2 t)}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d x}{d t} = [\frac{3 a + 3 a t^{2} - 6 a t^{2}}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d x}{d t} = [\frac{3 a - 3 a t^{2}}{{(1 - t^{2})}^{2}}] \Rightarrow \frac{d x}{d t} = \frac{3 a (1 - t^{2})}{{(1 + t^{2})}^{2}} . . . (i) and, y = \frac{3 a t^{2}}{1 + t^{2}}$
Differentiating it with respect to t,

$\frac{d x}{d t} = [\frac{(1 + t^{2}) \frac{d}{d t} (3 a t^{2}) - 3 a t^{2} \frac{d}{d t} (1 + t^{2})}{{(1 + t^{2})}^{2}}] [using quotient rule] \Rightarrow \frac{d x}{d t} = [\frac{(1 + t^{2}) (6 a t) - 3 a t^{2} (2 t)}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d x}{d t} = [\frac{6 a t + 6 a t^{3} - 6 a t^{3}}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d x}{d t} = \frac{6 a t}{{(1 + t^{2})}^{2}} . . . (i i) Dividing equation (ii) by (i), \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{6 a t}{{(1 + t^{2})}^{2}} \times \frac{{(1 + t^{2})}^{2}}{3 a (1 - t^{2})} = \frac{2 t}{1 - t^{2}}$

Page No 10.103:

Question 9:

Find $\frac{d y}{d x}$ , when

$x = a (\cos θ + θ \sin θ) and y = a (\sin θ - θ \cos θ)$

Answer:

$We have, x = a (\cos θ + θ \sin θ) a n d y = a (\sin θ - θ \cos θ) \Rightarrow \frac{d x}{d θ} = a [\frac{d}{d θ} \cos θ + \frac{d}{d θ} (θ \sin θ)] and \frac{d y}{d θ} = a [\frac{d}{d θ} (\sin θ) - \frac{d}{d θ} (θ \cos θ)] \Rightarrow \frac{d x}{d θ} = a [- \sin θ + θ \frac{d}{d θ} (\sin θ) + \sin θ \frac{d}{d θ} (θ)] and \frac{d y}{d θ} = a [\cos θ - \{θ \frac{d}{d θ} (\cos θ) + \cos θ \frac{d}{d θ} (θ)\}] \Rightarrow \frac{d x}{d θ} = a [- \sin θ + θ \cos θ] and \frac{d y}{d θ} = a [\cos θ + θ \sin θ - \cos θ] \Rightarrow \frac{d x}{d θ} = a θ \cos θ and \frac{d y}{d θ} = a θ \sin θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a θ \sin θ}{a θ \cos θ} = \tan θ$

Page No 10.103:

Question 10:

Find $\frac{d y}{d x}$ , when

$x = e^{θ} (θ + \frac{1}{θ}) and y = e^{- θ} (θ - \frac{1}{θ})$

Answer:

$We have, x = e^{θ} (θ + \frac{1}{θ})$
Differentiating it with respect to $θ$ ,
$\frac{d x}{d θ} = e^{θ} \frac{d}{d θ} (θ + \frac{1}{θ}) + (θ + \frac{1}{θ}) \frac{d}{d θ} (e^{θ}) [using product rule] \Rightarrow \frac{d x}{d θ} = e^{θ} (1 - \frac{1}{θ^{2}}) + (\frac{θ^{2} + 1}{θ}) e^{θ} \Rightarrow \frac{d x}{d θ} = e^{θ} (1 - \frac{1}{θ^{2}} + \frac{θ^{2} + 1}{θ}) \Rightarrow \frac{d x}{d θ} = e^{θ} (\frac{θ^{2} - 1 + θ^{3} + θ}{θ^{2}}) \Rightarrow \frac{d x}{d θ} = \frac{e^{θ} (θ^{3} + θ^{2} + θ - 1)}{θ^{2}} . . . (i) and, y = e^{θ} (θ - \frac{1}{θ})$
Differentiating it with respect to $θ$ using chain rule,

$\frac{d y}{d θ} = e^{- θ} \frac{d}{d θ} (θ - \frac{1}{θ}) + (θ - \frac{1}{θ}) \frac{d}{d θ} (e^{- θ}) [using product rule] \Rightarrow \frac{d y}{d θ} = e^{- θ} (1 + \frac{1}{θ^{2}}) + (θ - \frac{1}{θ}) e^{θ} \frac{d}{d θ} (- θ) \Rightarrow \frac{d y}{d θ} = e^{- θ} (1 + \frac{1}{θ^{2}}) + (θ - \frac{1}{θ}) e^{- θ} (- 1) \Rightarrow \frac{d y}{d θ} = e^{- θ} (1 + \frac{1}{θ^{2}} - θ + \frac{1}{θ}) \Rightarrow \frac{d y}{d θ} = e^{- θ} (\frac{θ^{2} + 1 - θ^{3} + θ}{θ^{2}}) \Rightarrow \frac{d y}{d θ} = \frac{e^{- θ} (- θ^{3} + θ^{2} + θ + 1)}{θ^{2}} . . . (i i) Dividing equation (ii) by (i), \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = e^{- θ} (\frac{θ^{2} - θ^{3} + θ + 1}{θ^{2}}) \times \frac{θ^{2}}{e^{θ} (θ^{3} + θ^{2} + θ - 1)} = e^{- 2 θ} (\frac{θ^{2} - θ^{3} + θ + 1}{θ^{3} + θ^{2} + θ - 1})$

Page No 10.103:

Question 11:

Find $\frac{d y}{d x}$ , when

$x = \frac{2 t}{1 + t^{2}} and y = \frac{1 - t^{2}}{1 + t^{2}}$

Answer:

$We have, x = \frac{2 t}{1 + t^{2}}$

$\Rightarrow \frac{d x}{d t} = [\frac{(1 + t^{2}) \frac{d}{d t} (2 t) - 2 t \frac{d}{d t} (1 + t^{2})}{{(1 + t^{2})}^{2}}] [using quotient rule] \Rightarrow \frac{d x}{d t} = [\frac{(1 + t^{2}) (2) - 2 t (2 t)}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d x}{d t} = [\frac{2 + 2 t^{2} - 4 t^{2}}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d x}{d t} = [\frac{2 - 2 t^{2}}{{(1 + t^{2})}^{2}}] . . . (i) and, y = \frac{1 - t^{2}}{1 + t^{2}}$

$\Rightarrow \frac{d y}{d t} = [\frac{(1 + t^{2}) \frac{d}{d t} (1 - t^{2}) - (1 - t^{2}) \frac{d}{d t} (1 + t^{2})}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d y}{d t} = [\frac{(1 + t^{2}) (- 2 t) - (1 - t^{2}) (2 t)}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d y}{d t} = [\frac{- 4 t}{{(1 + t^{2})}^{2}}] . . . (ii) Dividing equation (ii) by (i), we get, \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- 4 t}{{(1 + t^{2})}^{2}} \times \frac{{(1 + t^{2})}^{2}}{2 (1 - t^{2})} \Rightarrow \frac{d y}{d x} = \frac{- 2 t}{1 - t^{2}} \Rightarrow \frac{d y}{d x} = - \frac{x}{y} [∵ \frac{x}{y} = \frac{2 t}{1 + t^{2}} \times \frac{1 + t^{2}}{1 - t^{2}} = \frac{2 t}{1 - t^{2}}]$

Page No 10.103:

Question 12:

Find $\frac{d y}{d x}$ , when

$x = \cos^{- 1} \frac{1}{\sqrt{1 + t^{2}}} and y = \sin^{- 1} \frac{t}{\sqrt{1 + t^{2}}}, t \in R$

Answer:

$We have, x = \cos^{- 1} (\frac{1}{\sqrt{1 + t^{2}}})$

$\Rightarrow \frac{d x}{d t} = \frac{- 1}{\sqrt{1 - {(\frac{1}{\sqrt{1 + t^{2}}})}^{2}}} \frac{d}{d t} (\frac{1}{\sqrt{1 + t^{2}}}) \Rightarrow \frac{d x}{d t} = \frac{- 1}{\sqrt{1 - \frac{1}{(1 + t^{2})}}} \{\frac{- 1}{2 {(1 + t^{2})}^{\frac{3}{2}}}\} \frac{d}{d t} (1 + t^{2}) \Rightarrow \frac{d x}{d t} = \frac{{(1 + t^{2})}^{\frac{1}{2}}}{\sqrt{1 + t^{2} - 1}} \times \frac{1}{2 {(1 + t^{2})}^{\frac{3}{2}}} (2 t) \Rightarrow \frac{d x}{d t} = \frac{t}{\sqrt{t^{2}} \times (1 + t^{2})} \Rightarrow \frac{d x}{d t} = \frac{1}{1 + t^{2}} . . . (i) Now, y = \sin^{- 1} (\frac{1}{\sqrt{1 + t^{2}}})$

$\Rightarrow \frac{d y}{d t} = \frac{1}{\sqrt{1 - {(\frac{1}{\sqrt{1 + t^{2}}})}^{2}}} \frac{d}{d t} (\frac{1}{\sqrt{1 + t^{2}}}) \Rightarrow \frac{d y}{d t} = \frac{1}{\sqrt{1 - \frac{1}{(1 + t^{2})}}} \{\frac{- 1}{2 {(1 + t^{2})}^{\frac{3}{2}}}\} \frac{d}{d t} (1 + t^{2}) \Rightarrow \frac{d x}{d t} = \frac{{(1 + t^{2})}^{\frac{1}{2}}}{\sqrt{1 + t^{2} - 1}} \times \frac{- 1}{2 {(1 + t^{2})}^{\frac{3}{2}}} (2 t) \Rightarrow \frac{d x}{d t} = \frac{- 1}{2 \sqrt{t^{2}} \times (1 + t^{2})} (2 t) \Rightarrow \frac{d x}{d t} = \frac{- 1}{1 + t^{2}} . . . (ii) Dividing equation (ii) by (i), \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{1}{(1 + t^{2})} \times \frac{(1 + t^{2})}{- 1} \Rightarrow \frac{d y}{d x} = - 1$

Page No 10.103:

Question 13:

Find $\frac{d y}{d x}$ , when

$x = \frac{1 - t^{2}}{1 + t^{2}} and y = \frac{2 t}{1 + t^{2}}$

Answer:

$We have, y = \frac{2 t}{1 + t^{2}}$

$\Rightarrow \frac{d y}{d t} = [\frac{(1 + t^{2}) \frac{d}{d t} (2 t) - 2 t \frac{d}{d t} (1 + t^{2})}{{(1 + t^{2})}^{2}}] [using quotient rule] \Rightarrow \frac{d y}{d t} = [\frac{(1 + t^{2}) (2) - 2 t (2 t)}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d y}{d t} = [\frac{2 + 2 t^{2} - 4 t^{2}}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d y}{d t} = [\frac{2 - 2 t^{2}}{{(1 + t^{2})}^{2}}] . . . (i) and, x = \frac{1 - t^{2}}{1 + t^{2}}$

$\Rightarrow \frac{d x}{d t} = [\frac{(1 + t^{2}) \frac{d}{d t} (1 - t^{2}) - (1 - t^{2}) \frac{d}{d t} (1 + t^{2})}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d x}{d t} = [\frac{(1 + t^{2}) (- 2 t) - (1 - t^{2}) (2 t)}{{(1 + t^{2})}^{2}}] \Rightarrow \frac{d x}{d t} = [\frac{- 4 t}{{(1 + t^{2})}^{2}}] . . . (ii) Dividing equation (i) by (ii), we get, \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 (1 - t^{2})}{{(1 + t^{2})}^{2}} \times \frac{{(1 + t^{2})}^{2}}{- 4 t} \Rightarrow \frac{d y}{d x} = \frac{2 (1 - t^{2})}{- 4 t} \Rightarrow \frac{d y}{d x} = \frac{t^{2} - 1}{2 t}$

Page No 10.103:

Question 14:

If $x = 2 \cos θ - \cos 2 θ and y = 2 \sin θ - \sin 2 θ$ , prove that $\frac{d y}{d x} = \tan (\frac{3 θ}{2})$

Answer:

$We have, x = 2 \cos θ - \cos 2 θ$

$\Rightarrow \frac{d x}{d θ} = 2 (- \sin θ) - (- \sin 2 θ) \frac{d}{d θ} (2 θ) \Rightarrow \frac{d x}{d θ} = - 2 \sin θ + 2 \sin 2 θ \Rightarrow \frac{d x}{d θ} = 2 (\sin 2 θ - \sin θ) . . . (i) and, y = 2 \sin θ - \sin 2 θ$

$\Rightarrow \frac{d y}{d θ} = 2 \cos θ - \cos 2 θ \frac{d}{d θ} (2 θ) \Rightarrow \frac{d y}{d θ} = 2 \cos θ - \cos 2 θ (2) \Rightarrow \frac{d y}{d θ} = 2 \cos θ - 2 \cos 2 θ \Rightarrow \frac{d y}{d θ} = 2 (\cos θ - \cos 2 θ) . . . (ii) Dividing equation (ii) by equation (i), \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{2 (\cos θ - \cos 2 θ)}{2 (\sin 2 θ - \sin θ)} \Rightarrow \frac{d y}{d x} = \frac{\cos θ - \cos 2 θ}{\sin 2 θ - \sin θ} \Rightarrow \frac{d y}{d x} = \frac{- 2 \sin (\frac{θ + 2 θ}{2}) \sin (\frac{θ - 2 θ}{2})}{2 \cos (\frac{2 θ + θ}{2}) \sin (\frac{2 θ - θ}{2})} [\begin{matrix} ∵ \sin A - \sin B = 2 \cos (\frac{A + B}{2}) \sin (\frac{A - B}{2}) \\ and \cos A - \cos B = - 2 \sin (\frac{A + B}{2}) \sin (\frac{A - B}{2}) \end{matrix}] \Rightarrow \frac{d y}{d x} = \frac{- \sin (\frac{3 θ}{2}) \sin (\frac{- θ}{2})}{\cos (\frac{3 θ}{2}) \sin (\frac{θ}{2})} \Rightarrow \frac{d y}{d x} = \frac{- \sin (\frac{3 θ}{2}) (- \sin \frac{θ}{2})}{\cos (\frac{3 θ}{2}) \sin (\frac{θ}{2})} \Rightarrow \frac{d y}{d x} = \frac{\sin (\frac{3 θ}{2})}{\cos (\frac{3 θ}{2})} \Rightarrow \frac{d y}{d x} = \tan (\frac{3 θ}{2})$

Page No 10.103:

Question 15:

If $x = e^{\cos 2 t} and y = e^{\sin 2 t},$ prove that $\frac{d y}{d x} = - \frac{y \log x}{x \log y}$

Answer:

$We have, x = e^{\cos 2 t} and y = e^{\sin 2 t}$

$\Rightarrow \frac{d x}{d t} = \frac{d}{d t} (e^{\cos 2 t}) and \frac{d y}{d t} = \frac{d}{d t} (e^{\sin 2 t}) \Rightarrow \frac{d x}{d t} = e^{\cos 2 t} \frac{d}{d t} (\cos 2 t) and \frac{d y}{d t} = e^{\sin 2 t} \frac{d}{d t} (\sin 2 t) \Rightarrow \frac{d x}{d t} = e^{\cos 2 t} (- \sin 2 t) \frac{d}{d t} (2 t) and \frac{d y}{d t} = e^{\sin 2 t} (\cos 2 t) \frac{d}{d t} (2 t) \Rightarrow \frac{d x}{d t} = - 2 \sin 2 t e^{\cos 2 t} and \frac{d y}{d t} = 2 \cos 2 t e^{\sin 2 t} ∵ \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 \cos 2 t e^{\sin 2 t}}{- 2 \sin 2 t e^{\cos 2 t}} \Rightarrow \frac{d y}{d x} = - \frac{y \log x}{x \log y} [\begin{matrix} ∵ x = e^{\cos 2 t} \Rightarrow \log x = \cos 2 t \\ y = e^{\sin 2 t} \Rightarrow \log y = \sin 2 t \end{matrix}]$

Page No 10.103:

Question 16:

If $x = \cos t and y = \sin t,$ prove that $\frac{d y}{d x} = \frac{1}{\sqrt{3}} at t = \frac{2 π}{3}$

Answer:

$We have, x = \cos t and y = \sin t$

$\Rightarrow \frac{d x}{d t} = \frac{d}{d t} (\cos t) and \frac{d y}{d t} = \frac{d}{d t} (\sin t) \Rightarrow \frac{d x}{d t} = - \sin t and \frac{d y}{d t} = \cos t ∴ \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\cos t}{- \sin t} = - \cot t Now, {(\frac{d y}{d x})}_{t = \frac{2 π}{3}} = - \cot (\frac{2 π}{3}) = \frac{1}{\sqrt{3}}$

Page No 10.103:

Question 17:

If $x = a (t + \frac{1}{t}) and y = a (t - \frac{1}{t})$ , prove that $\frac{d y}{d x} = \frac{x}{y}$

Answer:

$We have, x = a (t + \frac{1}{t}) and y = a (t - \frac{1}{t})$

$\Rightarrow \frac{d x}{d t} = a \frac{d}{d t} (t + \frac{1}{t}) and \frac{d y}{d t} = a \frac{d}{d t} (t - \frac{1}{t}) \Rightarrow \frac{d x}{d t} = a (1 - \frac{1}{t^{2}}) and \frac{d y}{d t} = a (1 + \frac{1}{t^{2}}) \Rightarrow \frac{d x}{d t} = a (\frac{t^{2} - 1}{t^{2}}) and \frac{d y}{d t} = a (\frac{t^{2} + 1}{t^{2}}) ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a (t^{2} + 1)}{t^{2}} \times \frac{t^{2}}{a (t^{2} - 1)} \Rightarrow \frac{d y}{d x} = \frac{a (t^{2} + 1)}{t} \times \frac{t}{a (t^{2} - 1)} \Rightarrow \frac{d y}{d x} = a (t + \frac{1}{t}) \times \frac{1}{a (t - \frac{1}{t})} \Rightarrow \frac{d y}{d x} = \frac{x}{y}$

Page No 10.103:

Question 18:

If $x = \sin^{- 1} (\frac{2 t}{1 + t^{2}}) and y = \tan^{- 1} (\frac{2 t}{1 - t^{2}}), - 1 < t < 1$ , prove that $\frac{d y}{d x} = 1$

Answer:

$We have, x = \sin^{- 1} (\frac{2 t}{1 + t^{2}}) Put t = \tan θ \Rightarrow - 1 < \tan θ < 1 \Rightarrow - \frac{π}{4} < θ < \frac{π}{4} \Rightarrow - \frac{π}{2} < 2 θ < \frac{π}{2} ∴ x = \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ}) \Rightarrow x = \sin^{- 1} (\sin 2 θ) \Rightarrow x = 2 θ [∵ - \frac{π}{2} < 2 θ < \frac{π}{2}] \Rightarrow x = 2 (\tan^{- 1} t) [∵ t = \sin θ]$

$\Rightarrow \frac{d x}{d t} = \frac{2}{1 + t^{2}} . . . (i) Now, y = \tan^{- 1} (\frac{2 t}{1 - t^{2}}) put t = \tan θ \Rightarrow y = \tan^{- 1} (\frac{2 \tan θ}{1 - \tan^{2} θ}) \Rightarrow y = \tan^{- 1} (\tan 2 θ) \Rightarrow y = 2 θ [∵ - \frac{π}{2} < 2 θ < \frac{π}{2}] \Rightarrow y = 2 \tan^{- 1} t [∵ t = \tan θ]$

$\Rightarrow \frac{d y}{d t} = \frac{2}{1 + t^{2}} . . . (i i) Dividing equation (ii) by (i), \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2}{1 + t^{2}} \times \frac{1 + t^{2}}{2} \Rightarrow \frac{d y}{d x} = 1$

Page No 10.103:

Question 19:

If $x = \frac{\sin^{3} t}{\sqrt{\cos 2 t}}, y = \frac{\cos^{3} t}{\sqrt{\cos t 2 t}}$ , find $\frac{d y}{d x}$

Answer:

$We have, x = \frac{\sin^{3} t}{\sqrt{\cos 2 t}} and y = \frac{\cos^{3} t}{\sqrt{\cos 2 t}} \Rightarrow \frac{d x}{d t} = \frac{d}{d t} [\frac{\sin^{3} t}{\sqrt{\cos 2 t}}] \Rightarrow \frac{d x}{d t} = \frac{\sqrt{\cos 2 t} \frac{d}{d t} (\sin^{3} t) - \sin^{3} t \frac{d}{d t} \sqrt{\cos 2 t}}{\cos 2 t} [Using quotient rule] \Rightarrow \frac{d x}{d t} = \frac{\sqrt{\cos 2 t} (3 \sin^{2} t) \frac{d}{d t} (\sin t) - \sin^{3} t \times \frac{1}{2 \sqrt{\cos 2 t}} \frac{d}{d t} (\cos 2 t)}{\cos 2 t} \Rightarrow \frac{d x}{d t} = \frac{3 \sqrt{\cos 2 t} (\sin^{2} t \cos t) - \frac{\sin^{3} t}{2 \sqrt{\cos 2 t}} (- 2 \sin 2 t)}{\cos 2 t} \Rightarrow \frac{d x}{d t} = \frac{3 \cos 2 t \sin^{2} t \cos t + \sin^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} Now, \frac{d y}{d t} = \frac{d}{d t} [\frac{\cos^{3} t}{\sqrt{\cos 2 t}}] \Rightarrow \frac{d y}{d t} = \frac{\sqrt{\cos 2 t} \frac{d}{d t} (\cos^{3} t) - \cos^{3} t \frac{d}{d t} \sqrt{\cos 2 t}}{\cos 2 t} [Using quotient rule] \Rightarrow \frac{d y}{d t} = \frac{\sqrt{\cos 2 t} (3 \cos^{2} t) \frac{d}{d t} (\cos t) - \cos^{3} t \times \frac{1}{2 \sqrt{\cos 2 t}} \frac{d}{d t} (\cos 2 t)}{\cos 2 t} \Rightarrow \frac{d y}{d t} = \frac{3 \sqrt{\cos 2 t} \cos^{2} t (- \sin t) - \frac{\cos^{3} t}{2 \sqrt{\cos 2 t}} (- 2 \sin 2 t)}{\cos 2 t} \Rightarrow \frac{d y}{d t} = \frac{- 3 \cos 2 t \cos^{2} t \sin t + \cos^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- 3 \cos 2 t \cos^{2} t \sin t + \cos^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \times \frac{\cos 2 t \sqrt{\cos 2 t}}{3 \cos 2 t \sin^{2} t \cos t + \sin^{3} t \sin 2 t} \Rightarrow \frac{d y}{d x} = \frac{- 3 \cos 2 t \cos^{2} t \sin t + \cos^{3} t \sin 2 t}{3 \cos 2 t \sin^{2} t \cos t + \sin^{3} t \sin 2 t} \Rightarrow \frac{d y}{d x} = \frac{\sin t \cos t [- 3 \cos 2 t \cos t + 2 \cos^{3} t]}{\sin t \cos t [3 \cos 2 t \sin t + 2 \sin^{3} t]} \Rightarrow \frac{d y}{d x} = \frac{[- 3 (2 \cos^{2} t - 1) \cos t + 2 \cos^{3} t]}{[3 (1 - 2 \sin^{2} t) \sin t + 2 \sin^{3} t]} [\begin{matrix} \cos 2 t = 2 \cos^{2} t - 1 \\ \cos 2 t = 1 - 2 \sin^{2} t \end{matrix}] \Rightarrow \frac{d y}{d x} = \frac{- 4 \cos^{3} t + 3 \cos t}{3 \sin t - 4 \sin^{3} t} \Rightarrow \frac{d y}{d x} = \frac{- \cos 3 t}{\sin 3 t} [\begin{matrix} \cos 3 t = 4 \cos^{3} t - 3 \cos t \\ \sin 3 t = 3 \sin t - 4 \sin^{3} t \end{matrix}] ∴ \frac{d y}{d x} = - \cot 3 t$

Page No 10.103:

Question 20:

If $x = {(t + \frac{1}{t})}^{a}, y = a^{t + \frac{1}{t}}, find \frac{d y}{d x}$

Answer:

$We have, x = {(t + \frac{1}{t})}^{a}$

$\Rightarrow \frac{d x}{d t} = \frac{d}{d t} [{(t + \frac{1}{t})}^{a}] \Rightarrow \frac{d x}{d t} = a {(t + \frac{1}{t})}^{a - 1} \frac{d}{d t} (t + \frac{1}{t}) \Rightarrow \frac{d x}{d t} = a {(t + \frac{1}{t})}^{a - 1} (1 - \frac{1}{t^{2}}) . . . (i) and, y = a^{(t + \frac{1}{t})}$

$\Rightarrow \frac{d y}{d t} = \frac{d}{d t} [a^{(t + \frac{1}{t})}] \Rightarrow \frac{d y}{d t} = a^{(t + \frac{1}{t})} \times \log a \frac{d}{d t} (t + \frac{t}{t}) \Rightarrow \frac{d y}{d t} = a^{(t + \frac{1}{t})} \times \log a (1 - \frac{1}{t^{2}}) . . . (i i) Dividing equation (ii) by (i), \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a^{(t + \frac{1}{t})} \times \log a (1 - \frac{1}{t^{2}})}{a {(t + \frac{1}{t})}^{a - 1} (1 - \frac{1}{t^{2}})} \Rightarrow \frac{d y}{d x} = \frac{a^{(t + \frac{1}{t})} \times \log a}{a {(t + \frac{1}{t})}^{a - 1}}$

Page No 10.103:

Question 21:

If $x = a (\frac{1 + t^{2}}{1 - t^{2}}) and y = \frac{2 t}{1 - t^{2}}, find \frac{d y}{d x}$

Answer:

$We have, x = a (\frac{1 + t^{2}}{1 - t^{2}})$

$\Rightarrow \frac{d x}{d t} = a [\frac{(1 - t^{2}) \frac{d}{d t} (1 + t^{2}) - (1 + t^{2}) \frac{d}{d t} (1 - t^{2})}{{(1 - t^{2})}^{2}}] [Using quotient rule] \Rightarrow \frac{d x}{d t} = a [\frac{(1 - t^{2}) (2 t) - (1 + t^{2}) (- 2 t)}{{(1 - t^{2})}^{2}}] \Rightarrow \frac{d x}{d t} = a [\frac{2 t - 2 t^{3} + 2 t + 2 t^{3}}{{(1 - t^{2})}^{2}}] \Rightarrow \frac{d x}{d t} = \frac{4 a t}{{(1 - t^{2})}^{2}} . . . (i) and, y = \frac{2 t}{1 - t^{2}}$

$\Rightarrow \frac{d y}{d t} = 2 [\frac{(1 - t^{2}) \frac{d}{d t} (t) - t \frac{d}{d t} (1 - t^{2})}{{(1 - t^{2})}^{2}}] [Using quotient rule] \Rightarrow \frac{d y}{d t} = 2 [\frac{(1 - t^{2}) (1) - t (- 2 t)}{{(1 - t^{2})}^{2}}] \Rightarrow \frac{d y}{d t} = 2 [\frac{1 - t^{2} + 2 t^{2}}{{(1 - t^{2})}^{2}}] \Rightarrow \frac{d y}{d t} = \frac{2 (1 + t^{2})}{{(1 - t^{2})}^{2}} . . . (ii) Dividing equation (ii) by (i), \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 (1 + t^{2})}{{(1 - t^{2})}^{2}} \times \frac{{(1 - t^{2})}^{2}}{4 a t} \Rightarrow \frac{d y}{d x} = \frac{(1 + t^{2})}{2 a t}$

Page No 10.104:

Question 22:

If $x = 10 (t - \sin t), y = 12 (1 - \cos t), find \frac{d y}{d x} .$

Answer:

$We have, x = 10 (t - \sin t) and y = 12 (1 - \cos t) \Rightarrow \frac{d x}{d t} = \frac{d}{d t} [10 (t - \sin t)] and \frac{d y}{d t} = \frac{d}{d t} [12 (1 - \cos t)] \Rightarrow \frac{d x}{d t} = 10 \frac{d}{d t} (t - \sin t) and \frac{d y}{d t} = 12 \frac{d}{d t} (1 - \cos t) \Rightarrow \frac{d x}{d t} = 10 (1 - \cos t) and \frac{d y}{d t} = 12 [0 - (- \sin t)] = 12 \sin t ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{12 \sin t}{10 (1 - \cos t)} \Rightarrow \frac{d y}{d x} = \frac{12 \times 2 \sin \frac{t}{2} \cos \frac{t}{2}}{10 \times 2 \sin^{2} \frac{t}{2}} \Rightarrow \frac{d y}{d x} = \frac{6}{5} \cot \frac{t}{2}$

Page No 10.104:

Question 23:

If $x = a (θ - \sin θ) and, y = a (1 + \cos θ), find \frac{d y}{d x} at θ = \frac{π}{3} .$

Answer:

$We have, x = a (θ - \sin θ) a n d y = a (1 + \cos θ) \Rightarrow \frac{d x}{d θ} = \frac{d}{d θ} [a (θ - \sin θ)] and \frac{d y}{d θ} = \frac{d}{d θ} [a (1 + \cos θ)] \Rightarrow \frac{d x}{d θ} = a (1 - \cos θ) and \frac{d y}{d θ} = a (- \sin θ) ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{- a \sin θ}{a (1 - \cos θ)} Now, {[\frac{d y}{d x}]}_{θ = \frac{π}{3}} = - \frac{\sin \frac{π}{3}}{1 - \cos \frac{π}{3}} = - \frac{\frac{\sqrt{3}}{2}}{1 - \frac{1}{2}} = - \sqrt{3}$

Page No 10.104:

Question 24:

If $x = a \sin 2 t (1 + \cos 2 t) and y = b \cos 2 t (1 - \cos 2 t)$ , show that at $t = \frac{π}{4}, \frac{d y}{d x} = \frac{b}{a}$ $t = \frac{π}{4}, \frac{d y}{d x} = \frac{b}{a}$

Answer:

$x = a \sin 2 t (1 + \cos 2 t) and y = b \cos 2 t (1 - \cos 2 t) \Rightarrow \frac{d x}{d t} = 2 a \cos 2 t (1 + \cos 2 t) + 2 a \sin 2 t (1 - \cos 2 t) and \frac{d y}{d t} = - 2 b \sin 2 t (1 - \cos 2 t) + 2 b \cos 2 t (1 + \cos 2 t) \Rightarrow \frac{d x}{d t} = 2 a (\cos 2 t + \cos^{2} 2 t + \sin 2 t - \sin 2 t \cos 2 t) and \frac{d y}{d t} = 2 b (- \sin 2 t + \sin 2 t \cos 2 t + \cos 2 t + \cos^{2} 2 t) ∴ \frac{d y}{d t} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- 2 b (- \sin 2 t + \sin 2 t \cos 2 t + \cos 2 t + \cos^{2} 2 t)}{2 a (\cos 2 t + \cos^{2} 2 t + \sin 2 t - \sin 2 t \cos 2 t)} \Rightarrow {(\frac{d y}{d t})}_{t = \frac{π}{4}} = \frac{- 2 b (- \sin \frac{π}{2} + \sin \frac{π}{2} \cos \frac{π}{2} + \cos \frac{π}{2} + \cos^{2} \frac{π}{2})}{2 a (\cos \frac{π}{2} + \cos^{2} \frac{π}{2} + \sin \frac{π}{2} - \sin \frac{π}{2} \cos \frac{π}{2})} = \frac{b}{a}$

Page No 10.104:

Question 25:

$If x = \cos t (3 - 2 \cos^{2} t), y = \sin t (3 - 2 \sin^{2} t) find the value of \frac{d y}{d x} at t = \frac{π}{4}$

Answer:

$x = \cos t (3 - 2 \cos^{2} t) and y = \sin t (3 - 2 \sin^{2} t) \Rightarrow \frac{d x}{d t} = - \sin t (3 - 2 \cos^{2} t) + \cos t (4 \cos t \sin t) and \frac{d y}{d t} = \cos t (3 - 2 \sin^{2} t) + \sin t (- 4 \sin t \cos t) \Rightarrow \frac{d x}{d t} = - 3 \sin t + 6 \sin t \cos^{2} t and \frac{d y}{d t} = 3 \cos t - 6 \sin^{2} t \cos t \Rightarrow \frac{d x}{d t} = - 3 \sin t (1 - 2 \cos^{2} t) and \frac{d y}{d t} = 3 \cos t (1 - 2 \sin^{2} t) \Rightarrow \frac{d x}{d t} = 3 \sin t \cos 2 t and \frac{d y}{d t} = 3 \cos t \cos 2 t ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{3 \cos t \cos 2 t}{3 \sin t \cos 2 t} = \cot t Now, {(\frac{d y}{d x})}_{t = \frac{π}{4}} = \cot \frac{π}{4} = 1$

Page No 10.104:

Question 26:

If $x = \frac{1 + \log t}{t^{2}}, y = \frac{3 + 2 \log t}{t}, find \frac{d y}{d x}$

Answer:

$x = \frac{1 + \log t}{t^{2}} and y = \frac{3 + 2 \log t}{t} \Rightarrow \frac{d x}{d t} = \frac{t - 2 t - 2 t \log t}{t^{4}} and \frac{d y}{d t} = \frac{2 - 3 - 2 \log t}{t^{2}} \Rightarrow \frac{d x}{d t} = \frac{- 1 - 2 \log t}{t^{3}} and \frac{d y}{d t} = \frac{- 1 - 2 \log t}{t^{2}} ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{- 1 - 2 \log t}{t^{2}}}{\frac{- 1 - 2 \log t}{t^{3}}} = t$

Page No 10.104:

Question 27:

$\sin x = \frac{2 t}{1 + t^{2}}, \tan y = \frac{2 t}{1 - t^{2}}, find \frac{d y}{d x}$

Answer:

$\sin x = \frac{2 t}{1 + t^{2}} and \tan y = \frac{2 t}{1 - t^{2}} \Rightarrow x = \sin^{- 1} \frac{2 t}{1 + t^{2}} and y = \tan^{- 1} \frac{2 t}{1 - t^{2}} \Rightarrow x = 2 \tan^{- 1} t and y = 2 \tan^{- 1} t \Rightarrow \frac{d x}{d t} = \frac{2 t}{1 + t^{2}} and \frac{d y}{d t} = \frac{2 t}{1 + t^{2}} ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{2 t}{1 + t^{2}}}{\frac{2 t}{1 + t^{2}}} = 1$

Page No 10.104:

Question 28:

Write the derivative of sinx with respect to cosx

Answer:

$Let u = \sin x and v = \cos x \Rightarrow \frac{d u}{d x} = \cos x and \frac{d v}{d x} = - \sin x ∴ \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\cos x}{- \sin x} \Rightarrow \frac{d u}{d v} = - \cot x$

Page No 10.104:

Question 29:

If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find $\frac{d y}{d x}$ when $θ = \frac{π}{3}$ .

Answer:

Given values are:
$x = a (2 θ - \sin 2 θ) and y = a (1 - \cos 2 θ)$

Applying parametric differentiation

$\frac{d x}{d θ}$ = 2a − 2acos2 $θ$

$\frac{d y}{d θ}$ = 0 + 2asin2 $θ$

$\frac{d y}{d x}$ = $\frac{d y}{d θ} \times \frac{d θ}{d x} = \frac{\sin 2 θ}{1 - \cos 2 θ}$

Now putting the value of $θ$ = $\frac{π}{3}$
${\frac{d y}{d x}}_{θ = \frac{π}{3}} = \frac{\sin 2 (\frac{π}{3})}{1 - \cos 2 (\frac{π}{3})} = \frac{\frac{\sqrt{3}}{2}}{1 + \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{1}{\sqrt{3}}$
So, $\frac{d y}{d x}$ is $\frac{1}{\sqrt{3}}$ at $θ = \frac{π}{3}$ .

Page No 10.112:

Question 1:

Differentiate x² with respect to x³

Answer:

$Let u = x^{2} and v = x^{3} \Rightarrow \frac{d u}{d x} = 2 x and \frac{d v}{d x} = 3 x^{2} ∴ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{2 x}{3 x^{2}} = \frac{2}{3 x}$

Page No 10.112:

Question 2:

Differentiate log (1 + x²) with respect to tan⁻¹ x

Answer:

$Let u = \log (1 + x^{2}) and v = \tan^{- 1} x \Rightarrow \frac{d u}{d x} = \frac{1}{(1 + x^{2})} \frac{d}{d x} (1 + x^{2}) = \frac{2 x}{(1 + x^{2})} and \frac{d v}{d x} = \frac{1}{1 + x^{2}} ∴ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{2 x}{1 + x^{2}} \times \frac{1 + x^{2}}{1} = 2 x$

Page No 10.112:

Question 3:

Differentiate (log x)^x with respect to log x

Answer:

$Let u = {(\log x)}^{x}$
Taking log on both sides,
$\log u = \log {(\log x)}^{x} \Rightarrow \log u = x \log (\log x)$

$\Rightarrow \frac{1}{u} \frac{d u}{d x} = x \frac{d}{d x} \{\log (\log x)\} + \log (\log x) \frac{d}{d x} (x) \Rightarrow \frac{1}{u} \frac{d u}{d x} = x (\frac{1}{\log x}) \frac{d}{d x} (\log x) + \log \log x (1) \Rightarrow \frac{d u}{d x} = u [\frac{x}{\log x} (\frac{1}{x}) + \log \log x] \Rightarrow \frac{d u}{d x} = {(\log x)}^{x} [\frac{1}{\log x} + \log \log x] . . (i) Again, let v = \log x \Rightarrow \frac{d v}{d x} = \frac{1}{x} . . . (ii) Dividing equation (i) by (ii), we get \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{{(\log x)}^{x} [\frac{1}{\log x} + \log \log x]}{\frac{1}{x}} \Rightarrow \frac{d u}{d v} = \frac{{(\log x)}^{x} [\frac{1 + \log x (\log \log x)}{\log x}]}{\frac{1}{x}} \Rightarrow \frac{d u}{d v} = x {(\log x)}^{x^{- 1}} (1 + \log x \times \log \log x)$

Page No 10.112:

Question 4:

Differentiate $\sin^{- 1} \sqrt{1 - x^{2}}$ with respect to $\cos^{- 1} x, if$
(i) $x \in (0, 1)$
(ii) $x \in (- 1, 0)$

Answer:

$(i) Let, u = \sin^{- 1} \sqrt{1 - x^{2}} Put x = \cos θ \Rightarrow u = \sin^{- 1} \sqrt{1 - \cos^{2} θ} \Rightarrow u = \sin^{- 1} (\sin θ) . . . (i) And, v = \cos^{- 1} x . . . (ii) Now, x \in (0, 1) \Rightarrow \cos θ \in (0, 1) \Rightarrow θ \in (0, \frac{π}{2}) So, from equation (i), u = θ [Since, \sin^{- 1} (\sin θ) = θ i f θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow u = \cos^{- 1} x [Since, \cos θ = x]$

Differentiating it with respect to x,

$\frac{d u}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} . . . (iii) from equation (ii), v = \cos^{- 1} x$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} . . . (iv) Dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{- 1}{\sqrt{1 - x^{2}}} \times \frac{\sqrt{1 - x^{2}}}{- 1} ∴ \frac{d u}{d x} = 1$

$(ii) Let, u = \sin^{- 1} \sqrt{1 - x^{2}} Put x = \cos θ \Rightarrow u = \sin^{- 1} \sqrt{1 - \cos^{2} θ} \Rightarrow u = \sin^{- 1} (\sin θ) . . . (i) And, v = \cos^{- 1} x . . . (ii) Now, x \in (- 1, 0) \Rightarrow \cos θ \in (- 1, 0) \Rightarrow θ \in (\frac{π}{2}, π) So, from equation (i), u = π - θ [Since, \sin^{- 1} (\sin θ) = π - θ if θ \in (\frac{π}{2}, \frac{3 π}{2})] \Rightarrow u = π - \cos^{- 1} x [Since, x = \cos θ]$

Differentiating it with respect to x,

$\frac{d u}{d x} = 0 - \frac{- 1}{\sqrt{1 - x^{2}}} \Rightarrow \frac{d u}{d x} = \frac{1}{\sqrt{1 - x^{2}}} . . . (iii) from equation (ii), v = \cos^{- 1} x$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} . . . (iv) Dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{1}{\sqrt{1 - x^{2}}} \times \frac{\sqrt{1 - x^{2}}}{- 1} ∴ \frac{d u}{d x} = - 1$

Page No 10.112:

Question 5:

Differentiate $\sin^{- 1} (4 x \sqrt{1 - 4 x^{2}})$ with respect to $\sqrt{1 - 4 x^{2}}$ , if

(i) $x \in (- \frac{1}{2 \sqrt{2}}, \frac{1}{\sqrt{2 \sqrt{2}}})$

(ii) $x \in (\frac{1}{2 \sqrt{2}}, \frac{1}{2})$

(iii) $x \in (- \frac{1}{2}, - \frac{1}{2 \sqrt{2}})$

Answer:

$(i) Let, u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) put 2 x = \cos θ \Rightarrow u = \sin^{- 1} (2 \times \cos θ \sqrt{1 - \cos^{2} θ}) \Rightarrow u = \sin^{- 1} (2 \cos θ \sin θ) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) Let, v = \sqrt{1 - 4 x^{2}} . . . (ii) Here, x \in (- \frac{1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}}) \Rightarrow 2 x \in (- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \Rightarrow θ \in (\frac{π}{4}, \frac{3 π}{4}) So, from equation (i), u = π - 2 θ [Since, \sin^{- 1} (sinθ) = π - θ, if θ \in (\frac{π}{2}, π)] \Rightarrow u = π - 2 \cos^{- 1} (2 x) [Since, 2 x = \cos θ]$

Differentiating it with respect to x,

$\frac{d u}{d x} = 0 - 2 (\frac{- 1}{\sqrt{1 - {(2 x)}^{2}}}) \frac{d}{d x} (2 x) \Rightarrow \frac{d u}{d x} = \frac{2}{\sqrt{1 - 4 x^{2}}} (2) \Rightarrow \frac{d u}{d x} = \frac{4}{\sqrt{1 - 4 x^{2}}} . . . (iii) from equation (ii) \frac{d v}{d x} = \frac{- 4 x}{\sqrt{1 - 4 x^{2}}} but, x \in (- \frac{1}{2}, - \frac{1}{2 \sqrt{2}}) \frac{d v}{d x} = \frac{- 4 (- x)}{\sqrt{1 - 4 {(- x)}^{2}}} \Rightarrow \frac{d v}{d x} = \frac{4 x}{\sqrt{1 - 4 x^{2}}} . . . (iv) Diferentiating equation (ii) with respect to x, \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - 4 x^{2}}} \frac{d}{d x} (1 - 4 x^{2}) \Rightarrow \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - 4 x^{2}}} (- 8 x) \Rightarrow \frac{d v}{d x} = \frac{- 4 x}{\sqrt{1 - 4 x^{2}}} . . . (v) Dividing equation (iii) by (v) \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{4}{\sqrt{1 - 4 x^{2}}} \times \frac{\sqrt{1 - 4 x^{2}}}{- 4 x} ∴ \frac{d u}{d v} = - \frac{1}{x}$

$(ii) Let, u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) put 2 x = \cos θ u = \sin^{- 1} (2 \times \cos θ \sqrt{1 - \cos^{2} θ}) \Rightarrow u = \sin^{- 1} (2 \cos θ \sin θ) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) Let, v = \sqrt{1 - 4 x^{2}} . . . (ii) Here, x \in (\frac{1}{2 \sqrt{2}}, \frac{1}{2}) \Rightarrow 2 x \in (\frac{1}{\sqrt{2}}, 1) \Rightarrow \cos θ \in (\frac{1}{\sqrt{2}}, 1) \Rightarrow θ \in (0, \frac{π}{4}) So, from equation (i), u = 2 θ [Since, \sin^{- 1} (sinθ) = θ, if θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow u = 2 \cos^{- 1} (2 x) [Since, 2 x = \cos θ]$

Differentiate it with respect to x,

$\frac{d u}{d x} = 2 (\frac{- 1}{\sqrt{1 - {(2 x)}^{2}}}) \frac{d}{d x} (2 x) \frac{d u}{d x} = \frac{- 2}{\sqrt{1 - 4 x^{2}}} (2) \frac{d u}{d x} = \frac{- 4}{\sqrt{1 - 4 x^{2}}} . . . (iii) Diferentiating equation (ii) with respect to x, \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - 4 x^{2}}} \frac{d}{d x} (1 - 4 x^{2}) \Rightarrow \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - 4 x^{2}}} (- 8 x) \Rightarrow \frac{d v}{d x} = \frac{- 4 x}{\sqrt{1 - 4 x^{2}}} . . . (iv) Dividing equation (iii) by (iv) \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{- 4}{\sqrt{1 - 4 x^{2}}} \times \frac{\sqrt{1 - 4 x^{2}}}{- 4 x} ∴ \frac{d u}{d v} = \frac{1}{x}$

$(iii) Let, u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) put, 2 x = \cos θ \Rightarrow u = \sin^{- 1} (2 \times \cos θ \sqrt{1 - \cos^{2} θ}) \Rightarrow u = \sin^{- 1} (2 \cos θ \sin θ) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) Let, v = \sqrt{1 - 4 x^{2}} . . . (ii) Here, x \in (- \frac{1}{2}, - \frac{1}{2 \sqrt{2}}) \Rightarrow 2 x \in (- 1, - \frac{1}{\sqrt{2}}) \Rightarrow θ \in (\frac{3 π}{4}, π) So, from equation (i), u = π - 2 θ [Since, \sin^{- 1} (sinθ) = π - θ, if θ \in (\frac{π}{2}, \frac{3 π}{2})] \Rightarrow u = π - 2 \cos^{- 1} (2 x) [Since, 2 x = \cos θ]$

Differentiate it with respect to x,

$\frac{d u}{d x} = 0 - 2 (\frac{- 1}{\sqrt{1 - {(2 x)}^{2}}}) \frac{d}{d x} (2 x) \Rightarrow \frac{d u}{d x} = \frac{2}{\sqrt{1 - 4 x^{2}}} (2) \Rightarrow \frac{d u}{d x} = \frac{4}{\sqrt{1 - 4 x^{2}}} . . . (iii) from equation (ii), \frac{d v}{d x} = \frac{- 4 x}{\sqrt{1 - 4 x^{2}}} but, x \in (- \frac{1}{2}, - \frac{1}{2 \sqrt{2}}) ∴ \frac{d v}{d x} = \frac{- 4 (- x)}{\sqrt{1 - 4 {(- x)}^{2}}} \Rightarrow \frac{d v}{d x} = \frac{4 x}{\sqrt{1 - 4 x^{2}}} . . . (iv) Dividing equation (iii) by (iv) \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{4}{\sqrt{1 - 4 x^{2}}} \times \frac{\sqrt{1 - 4 x^{2}}}{4 x} ∴ \frac{d u}{d v} = \frac{1}{x}$

Page No 10.112:

Question 6:

Differentiate $\tan^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x})$ with respect to $\sin^{- 1} (\frac{2 x}{1 + x^{2}})$ , if $- 1 < x < 1, x \neq 0 .$

Answer:

$Let, u = \tan^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x}) put x = \tan θ \Rightarrow u = \tan^{- 1} (\frac{\sqrt{1 + \tan^{2} θ} - 1}{\tan θ}) \Rightarrow u = \tan^{- 1} (\frac{s e c θ - 1}{\tan θ}) \Rightarrow u = \tan^{- 1} (\frac{1 - \cos θ}{\sin θ}) \Rightarrow u = \tan^{- 1} (\frac{2 \sin^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}) \Rightarrow u = \tan^{- 1} (\tan \frac{θ}{2}) . . . (i) And, v = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) \Rightarrow v = \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ}) \Rightarrow v = \sin^{- 1} (\sin 2 θ) . . . (ii) Here, - 1 < x < 1 \Rightarrow - 1 < \tan θ < 1 \Rightarrow - \frac{π}{4} < θ < \frac{π}{4} . . . (A) So, from equation (i), u = \frac{θ}{2} [Since, \tan^{- 1} (\tan θ) = θ, if θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow u = \frac{1}{2} \tan^{- 1} x [since, x = \tan θ]$

Differentiating it with respect to x,

$\frac{d u}{d x} = \frac{1}{2} (\frac{1}{1 + x^{2}}) \Rightarrow \frac{d u}{d x} = \frac{1}{2 (1 + x^{2})} . . . (i) Now, from equation (ii) and (A), v = 2 θ [Since, \sin^{- 1} (\sin θ) = θ, if θ \in [- \frac{π}{2}, \frac{π}{2}]] \Rightarrow v = 2 \tan^{- 1} x [Since, x = \tan θ]$

Differentiating it with respect to x,

$\frac{d v}{d x} = 2 (\frac{1}{1 + x^{2}}) . . . (iv) dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{1}{2 (1 + x^{2})} \times \frac{1 + x^{2}}{2} ∴ \frac{d u}{d v} = \frac{1}{4}$

Page No 10.112:

Question 7:

Differentiate $\sin^{- 1} (2 x \sqrt{1 - x^{2}})$ with respect to $\sec^{- 1} (\frac{1}{\sqrt{1 - x^{2}}})$ , if
(i) $x \in (0, \frac{1}{\sqrt{2}})$

(ii) $x \in (\frac{1}{\sqrt{2}}, 1)$

Answer:

$(i) Let, u = \sin^{- 1} (2 x \sqrt{1 - x^{2}}) Put x = \sin θ \Rightarrow u = \sin^{- 1} (2 \sin θ \sqrt{1 - \sin^{2} θ}) \Rightarrow u = \sin^{- 1} (2 \sin θ \cos θ) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) And, Let v = s e c^{- 1} (\frac{1}{\sqrt{1 - x^{2}}}) \Rightarrow v = s e c^{- 1} (\frac{1}{\sqrt{1 - \sin^{2} θ}}) \Rightarrow v = s e c^{- 1} (\frac{1}{\cos θ}) \Rightarrow v = s e c^{- 1} (s e c θ) \Rightarrow v = \cos^{- 1} (\frac{1}{\frac{1}{\cos θ}}) [Since, s e c^{- 1} x = \cos^{- 1} (\frac{1}{x})] \Rightarrow v = \cos^{- 1} (\cos θ) . . . (ii) Here, x \in (0, \frac{1}{\sqrt{2}}) \Rightarrow \sin θ \in (0, \frac{1}{\sqrt{2}}) \Rightarrow θ \in (0, \frac{π}{4}) So, from equation (i), u = 2 θ [Since, \sin^{- 1} (\sin θ) = θ, if θ \in (- \frac{π}{2}, \frac{π}{2})] Let, u = 2 \sin^{- 1} x [Since, x = \sin θ]$

Differentiating it with respect to x,

$\frac{d u}{d x} = 2 (\frac{1}{\sqrt{1 - x^{2}}}) \Rightarrow \frac{d u}{d x} = \frac{2}{\sqrt{1 - x^{2}}} . . . (iii) And, from equation (ii), v = θ [Since, \cos^{- 1} (\cos θ) = θ, if θ \in [0, π]] \Rightarrow v = \sin^{- 1} x [Since, x = \sin θ]$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{1}{\sqrt{1 - x^{2}}} . . . (iv) dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{2}{\sqrt{1 - x^{2}}} \times \frac{\sqrt{1 - x^{2}}}{1} ∴ \frac{d u}{d v} = 2$

$(ii) Let, u = \sin^{- 1} (2 x \sqrt{1 - x^{2}}) Put x = \sin θ \Rightarrow u = \sin^{- 1} (2 \sin θ \sqrt{1 - \sin^{2} θ}) \Rightarrow u = \sin^{- 1} (2 \sin θ \cos θ) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) And, Let, v = s e c^{- 1} (\frac{1}{\sqrt{1 - x^{2}}}) \Rightarrow v = s e c^{- 1} (\frac{1}{\sqrt{1 - \sin^{2} θ}}) \Rightarrow v = s e c^{- 1} (\frac{1}{\cos θ}) \Rightarrow v = s e c^{- 1} (s e c θ) \Rightarrow v = \cos^{- 1} (\frac{1}{\frac{1}{\cos θ}}) [Since, s e c^{- 1} x = \cos^{- 1} (\frac{1}{x})] \Rightarrow v = \cos^{- 1} (\cos θ) . . . (ii) Here, x \in (\frac{1}{\sqrt{2}}, 1) \Rightarrow \sin θ \in (\frac{1}{\sqrt{2}}, 1) \Rightarrow θ \in (\frac{π}{4}, \frac{π}{2}) So, from equation (i), u = 2 θ [Since, \sin^{- 1} (\sin θ) = θ, if θ \in (- \frac{π}{2}, \frac{π}{2})] Let, u = 2 \sin^{- 1} x [Since, x = \sin θ]$

Differentiating it with respect to x,

$\frac{d u}{d x} = 2 (\frac{1}{\sqrt{1 - x^{2}}}) \Rightarrow \frac{d u}{d x} = \frac{2}{\sqrt{1 - x^{2}}} . . . (iii) And, from equation (ii), v = θ [Since, \cos^{- 1} (\cos θ) = θ, if θ \in [0, π]] \Rightarrow v = \sin^{- 1} x [Since, x = \sin θ]$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{1}{\sqrt{1 - x^{2}}} . . . (iv) dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{2}{\sqrt{1 - x^{2}}} \times \frac{\sqrt{1 - x^{2}}}{1} ∴ \frac{d u}{d v} = 2$

Page No 10.112:

Question 8:

Differentiate ${(\cos x)}^{\sin x}$ with respect to ${(\sin x)}^{\cos x}$ .

Answer:

$Let, u = {(\cos x)}^{\sin x}$

Taking log on both sides,

$\log u = \log {(\cos x)}^{\sin x} \Rightarrow \log u = \sin x \log (\cos x)$

Differentiating it with respect to x using chain rule,

$\frac{1}{u} \frac{d u}{d x} = \sin x \frac{d}{d x} (\log \cos x) + \log \cos x \frac{d}{d x} (\sin x) [using product rule] \Rightarrow \frac{1}{u} \frac{d u}{d x} = \sin x (\frac{1}{\cos x}) \frac{d}{d x} (\cos x) + \log \cos x (\cos x) \Rightarrow \frac{d u}{d x} = u [(\tan x) \times (- \sin x) + \log \log x (\cos x)] \Rightarrow \frac{d u}{d x} = {(\cos x)}^{\sin x} [\cos x \log \cos x - \sin x \tan x] . . . (i) Let, v = {(\sin x)}^{\cos x}$

Taking log on both sides,

$\log v = \log {(\sin x)}^{\cos x} \Rightarrow \log v = \cos x \log (\sin x)$

Differentiating it with respect to x using chain rule,

$\frac{1}{v} \frac{d v}{d x} = \cos x \frac{d}{d x} (\log \sin x) + logsin x \frac{d}{d x} (\cos x) [using product rule] \Rightarrow \frac{1}{v} \frac{d v}{d x} = \cos x (\frac{1}{\sin x}) \frac{d}{d x} (\sin x) + \log \sin x (- \sin x) \Rightarrow \frac{d v}{d x} = v [c o t x (\cos x) - \sin x \log \sin x] \Rightarrow \frac{d v}{d x} = {(\sin x)}^{\cos x} [c o t x (\cos x) - \sin x \log \sin x] dividing equation (i) by (ii), ∴ \frac{d u}{d v} = \frac{{(\cos x)}^{\sin x} [\cos x \log \cos x - \sin x \tan x]}{{(\sin x)}^{\cos x} [c o t x (\cos x) - \sin x \log \sin x]}$

Page No 10.112:

Question 9:

Differentiate $\sin^{- 1} (\frac{2 x}{1 + x^{2}})$ with respect to $\cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), if 0 < x < 1$

Answer:

$Let, u = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) Put x = \tan θ \Rightarrow u = \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ}) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) L e t v = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) \Rightarrow v = \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) \Rightarrow v = \cos^{- 1} (\cos 2 θ) . . . (ii) Here, 0 < x < 1 \Rightarrow 0 < \tan θ < 1 \Rightarrow 0 < θ < \frac{π}{4} So, from equation (i), u = 2 θ [Since, \sin^{- 1} (\sin θ) = θ, i f θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow u = 2 \tan^{- 1} x [Since, x = \tan θ]$

Differentiating it with respect to x,

$\frac{d u}{d x} = \frac{2}{1 + x^{2}} . . . (iii) from equation (ii), v = 2 θ [Since, \cos^{- 1} (\cos θ) = θ, if θ \in [0, π]] \Rightarrow v = 2 \tan^{- 1} x [Since, x = \tan θ]$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{2}{1 + x^{2}} . . . (iv) Dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{2}{1 + x^{2}} \times \frac{1 + x^{2}}{2} ∴ \frac{d u}{d v} = 1$

Page No 10.113:

Question 10:

Differentiate $\tan^{- 1} (\frac{1 + a x}{1 - a x})$ with respect to $\sqrt{1 + a^{2} x^{2}}$

Answer:

$Let, u = \tan^{- 1} (\frac{1 + a x}{1 - a x}) Put a x = \tan θ \Rightarrow u = \tan^{- 1} (\frac{1 + \tan θ}{1 - \tan θ}) \Rightarrow u = \tan^{- 1} (\frac{\tan \frac{π}{4} + \tan θ}{1 - \tan \frac{π}{4} \tan θ}) \Rightarrow u = \tan^{- 1} [\tan (\frac{π}{4} + θ)] \Rightarrow u = \frac{π}{4} + θ \Rightarrow u = \frac{π}{4} + \tan^{- 1} (a x) [Since, \tan θ = a x]$

Differentiating it with respect to x,

$\frac{d u}{d x} = 0 + \frac{1}{1 + {(a x)}^{2}} \frac{d}{d x} (a x) \Rightarrow \frac{d u}{d x} = \frac{a}{1 + a^{2} x^{2}} . . . (i) Now, Let, v = \sqrt{1 + a^{2} x^{2}}$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{1}{2 \sqrt{1 + a^{2} x^{2}}} \frac{d}{d x} (1 + a^{2} x^{2}) \Rightarrow \frac{d v}{d x} = \frac{1}{2 \sqrt{1 + a^{2} x^{2}}} (2 a^{2} x) \Rightarrow \frac{d v}{d x} = \frac{a^{2} x}{\sqrt{1 + a^{2} x^{2}}} . . . (ii) Dividing equation (i) by (ii), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{a}{1 + a^{2} x^{2}} \times \frac{\sqrt{1 + a^{2} x^{2}}}{a^{2} x} \frac{d u}{d v} = \frac{1}{a x \sqrt{1 + a^{2} x^{2}}}$

Page No 10.113:

Question 11:

Differentiate $\sin^{- 1} (2 x \sqrt{1 - x^{2}})$ with respect to $\tan^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}), if - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

Answer:

$Let, u = \sin^{- 1} (2 x \sqrt{1 - x^{2}}) Put x = \sin θ \Rightarrow u = \sin^{- 1} (2 \sin θ \sqrt{1 - \sin^{2} θ}) \Rightarrow u = \sin^{- 1} (2 \sin θ \cos θ) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) Let v = \tan^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}) \Rightarrow v = \tan^{- 1} (\frac{\sin θ}{\sqrt{1 - \sin^{2} θ}}) \Rightarrow v = \tan^{- 1} (\frac{\sin θ}{\cos θ}) \Rightarrow v = \tan^{- 1} (\tan θ) . . . (ii) Here, - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \Rightarrow - \frac{1}{\sqrt{2}} < \sin θ < \frac{1}{\sqrt{2}} \Rightarrow - \frac{π}{4} < θ < \frac{π}{4} So, from equation (i), u = 2 θ [Since, \sin^{- 1} (\sin θ) = θ, i f θ \in [- \frac{π}{2}, \frac{π}{2}]] \Rightarrow u = 2 \sin^{- 1} x [Since, x = \sin θ]$

Differentiating it with respect to x,

$\frac{d u}{d x} = \frac{2}{\sqrt{1 - x^{2}}} . . . (iii) from equation (ii), v = θ [Since, \tan^{- 1} (\tan θ) = θ, if θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow v = \sin^{- 1} x [Since, x = \sin θ]$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{1}{\sqrt{1 - x^{2}}} . . . (iv) Dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{2}{\sqrt{1 - x^{2}}} \times \frac{\sqrt{1 - x^{2}}}{1} ∴ \frac{d u}{d v} = 2$

Page No 10.113:

Question 12:

Differentiate $\tan^{- 1} (\frac{2 x}{1 - x^{2}})$ with respect to $\cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), if 0 < x < 1$

Answer:

$Let, u = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) Put x = \tan θ \Rightarrow u = \tan^{- 1} (\frac{2 \tan θ}{1 - \tan^{2} θ}) \Rightarrow u = \tan^{- 1} (\tan 2 θ) . . . (i) let, v = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) \Rightarrow v = \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) \Rightarrow v = \cos^{- 1} (\cos 2 θ) . . . (ii) Here, 0 < x < 1 \Rightarrow 0 < \tan θ < 1 \Rightarrow 0 < θ < \frac{π}{4} So, from equation (i), u = 2 θ [Since, \tan^{- 1} (\tan θ) = θ, if θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow u = 2 \tan^{- 1} x [Since, x = \tan θ]$

differentiating it with respect to x,

$\frac{d u}{d x} = \frac{2}{1 + x^{2}} . . . (iii) From equation (ii), v = θ [Since, \cos^{- 1} (\cos θ) = θ, if θ \in [0, π]] \Rightarrow v = 2 \tan^{- 1} x [Since, x = \tan θ]$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{2}{1 + x^{2}} . . . (iv) Dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{2}{1 + x^{2}} \times \frac{1 + x^{2}}{2} ∴ \frac{d u}{d v} = 1$

Page No 10.113:

Question 13:

Differentiate $\tan^{- 1} (\frac{x - 1}{x + 1})$ with respect to $\sin^{- 1} (3 x - 4 x^{3}), if - \frac{1}{2} < x < \frac{1}{2}$

Answer:

$Let, u = \tan^{- 1} (\frac{x - 1}{x + 1}) Put x = \tan θ \Rightarrow u = \tan^{- 1} (\frac{\tan θ - 1}{\tan θ + 1}) \Rightarrow u = \tan^{- 1} (\frac{\tan θ - \tan \frac{π}{4}}{1 + \tan θ \tan \frac{π}{4}}) \Rightarrow u = \tan^{- 1} [\tan (θ - \frac{π}{4})] . . . (i) Here, - \frac{1}{2} < x < \frac{1}{2} \Rightarrow - \frac{1}{2} < \tan θ < \frac{1}{2} \Rightarrow - \tan^{- 1} (\frac{1}{2}) < θ < \tan^{- 1} (\frac{1}{2}) So, from equation (i), u = θ - \frac{π}{4} [Since, \tan^{- 1} (\tan θ) = θ, i f θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow u = \tan^{- 1} x - \frac{π}{4} [Since, x = \tan θ]$

differentiating it with respect to x,

$\frac{d u}{d x} = \frac{1}{1 + x^{2}} - 0 \Rightarrow \frac{d u}{d x} = \frac{1}{1 + x^{2}} . . . (ii) And, Let, v = \sin^{- 1} (3 x - 4 x^{3}) Put x = \sin θ \Rightarrow v = \sin^{- 1} (3 \sin θ - 4 \sin^{3} θ) \Rightarrow v = \sin^{- 1} (\sin 3 θ) . . . (iii) Now, - \frac{1}{2} < x < \frac{1}{2} \Rightarrow - \frac{1}{2} < \sin θ < \frac{1}{2} \Rightarrow - \frac{1}{6} < θ < \frac{π}{6} So, from equation (iii), v = 3 θ [Since, \sin^{- 1} (\sin θ) = θ, i f θ \in [- \frac{π}{2}, \frac{π}{2}]] \Rightarrow v = 3 \sin^{- 1} x [Since, x = \sin θ]$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{3}{\sqrt{1 - x^{2}}} . . . (iv) Dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{1}{1 + x^{2}} \times \frac{\sqrt{1 - x^{2}}}{3} ∴ \frac{d u}{d v} = \frac{\sqrt{1 - x^{2}}}{3 (1 + x^{2})}$

Page No 10.113:

Question 14:

Differentiate $\tan^{- 1} (\frac{\cos x}{1 + \sin x})$ with respect to $\sec^{- 1} x$

Answer:

$Let, u = \tan^{- 1} (\frac{\cos x}{1 + \sin x}) \Rightarrow u = \tan^{- 1} [\tan (\frac{π}{4} - \frac{x}{2})] \Rightarrow u = \frac{π}{4} - \frac{x}{2}$

Differentiating it with respect to x,

$\frac{d u}{d x} = 0 - (\frac{1}{2}) \frac{d u}{d x} = - \frac{1}{2} . . . (i) Let, v = s e c^{- 1} x$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{1}{x \sqrt{x^{2} - 1}} . . . (ii) Dividing equation (i) by (ii), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = - \frac{1}{2} \times \frac{x \sqrt{x^{2} - 1}}{1} \frac{d u}{d v} = \frac{- x \sqrt{x^{2} - 1}}{2}$

Page No 10.113:

Question 15:

Differentiate $\sin^{- 1} (\frac{2 x}{1 + x^{2}})$ with respect to $\tan^{- 1} (\frac{2 x}{1 - x^{2}}), if - 1 < x < 1$

Answer:

$Let, u = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) Put x = \tan θ \Rightarrow θ = \tan^{- 1} x, \Rightarrow u = \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ}) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) Let, v = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) \Rightarrow v = \tan^{- 1} (\frac{2 \tan θ}{1 - \tan^{2} θ}) \Rightarrow v = \tan^{- 1} (\tan 2 θ) . . . (ii) Here, - 1 < x < 1 \Rightarrow - 1 < \tan θ < 1 \Rightarrow - \frac{π}{4} < \tan θ < \frac{π}{4} So, from equation (i), u = 2 θ [Since, \sin^{- 1} (\sin θ) = θ, if θ \in [- \frac{π}{2}, \frac{π}{2}]] \Rightarrow u = 2 \tan^{- 1} x$

Differentiating it with respect to x,

$\frac{d u}{d x} = \frac{2}{1 + x^{2}} . . . (iii) from equation (ii), v = 2 θ [Since, \tan^{- 1} (\tan θ) = θ, if θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow v = 2 \tan^{- 1} x$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{2}{1 + x^{2}} . . . (iv) Dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{2}{1 + x^{2}} \times \frac{1 + x^{2}}{2} ∴ \frac{d u}{d v} = 1$

Page No 10.113:

Question 16:

Differentiate $\cos^{- 1} (4 x^{3} - 3 x)$ with respect to $\tan^{- 1} (\frac{\sqrt{1 - x^{2}}}{x}), if \frac{1}{2} < x < 1$

Answer:

$Let, u = \cos^{- 1} (4 x^{3} - 3 x) Put, x = \cos θ \Rightarrow θ = \cos^{- 1} x Now, u = \cos^{- 1} (4 \cos^{3} θ - 3 \cos θ) \Rightarrow u = \cos^{- 1} (\cos 3 θ) . . . (i) Let, v = \tan^{- 1} (\frac{\sqrt{1 - x^{2}}}{x}) \Rightarrow v = \tan^{- 1} (\frac{\sqrt{1 - \cos^{2} θ}}{\cos θ}) \Rightarrow v = \tan^{- 1} (\frac{\sin θ}{\cos θ}) \Rightarrow v = \tan^{- 1} (\tan θ) . . . (ii) Here, \frac{1}{2} < x < 1 \Rightarrow \frac{1}{2} < \cos θ < 1 \Rightarrow 0 < θ < \frac{π}{3} So, from equation (i), u = 3 θ [Since, \cos^{- 1} (\cos θ) = θ, if θ \in [0, π]] \Rightarrow u = 3 \cos^{- 1} x$

Differentiating it with respect to x,

$\frac{d u}{d x} = \frac{- 3}{\sqrt{1 - x^{2}}} . . . (iii) From equation (ii), v = θ [Since, \tan^{- 1} (\tan θ) = θ, if θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow v = \cos^{- 1} x$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} . . . (iv) Dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = (\frac{- 3}{\sqrt{1 - x^{2}}}) (- \frac{\sqrt{1 - x^{2}}}{1}) ∴ \frac{d u}{d v} = 3$

Page No 10.113:

Question 17:

Differentiate $\tan^{- 1} (\frac{x}{\sqrt{1 - x^{2}}})$ with respect to $\sin^{- 1} (2 x \sqrt{1 - x^{2}}), if - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

Answer:

$Let, u = \tan^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}) Put x = \sin θ \Rightarrow θ = \sin^{- 1} x \Rightarrow u = \tan^{- 1} (\frac{\sin θ}{\sqrt{1 - \sin^{2} θ}}) \Rightarrow u = \tan^{- 1} (\frac{\sin θ}{\cos θ}) \Rightarrow u = \tan^{- 1} (\tan θ) . . . (i) And Let, v = \sin^{- 1} (2 x \sqrt{1 - x^{2}}) v = \sin^{- 1} (2 \sin θ \sqrt{1 - \sin^{2} θ}) v = \sin^{- 1} (2 \sin θ \cos θ) v = \sin^{- 1} (\sin 2 θ) . . . (ii) Here, - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \Rightarrow - \frac{1}{\sqrt{2}} < \sin θ < \frac{1}{\sqrt{2}} \Rightarrow - \frac{π}{4} < θ < \frac{π}{4} So, from equation (i), u = θ [Since, \tan^{- 1} (\tan θ) = θ, if θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow u = \sin^{- 1} x$

Differentiating it with respect to x,

$\frac{d u}{d x} = \frac{1}{\sqrt{1 - x^{2}}} . . . (iii) from equation (ii), v = 2 θ [Since, \sin^{- 1} (\sin θ) = θ, if θ \in [- \frac{π}{2}, \frac{π}{2}]] \Rightarrow v = 2 \sin^{- 1} x$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{2}{\sqrt{1 - x^{2}}} . . . (iv) Dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = (\frac{1}{\sqrt{1 - x^{2}}}) (\frac{\sqrt{1 - x^{2}}}{2}) ∴ \frac{d u}{d v} = \frac{1}{2}$

Page No 10.113:

Question 18:

Differentiate $\sin^{- 1} \sqrt{1 - x^{2}}$ with respect to $\cot^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}), if 0 < x < 1$

Answer:

$Let, u = \sin^{- 1} (\sqrt{1 - x^{2}}) Put x = \cos θ \Rightarrow θ = \cos^{- 1} x We get, u = \sin^{- 1} (\sin θ) . . . (i) Let, v = c o t^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}) \Rightarrow v = c o t^{- 1} (\frac{\cos θ}{\sqrt{1 - \cos^{2} θ}}) \Rightarrow v = c o t^{- 1} (\frac{\cos θ}{\sin θ}) \Rightarrow v = c o t^{- 1} (c o t θ) . . . (ii) Here, 0 < x < 1 \Rightarrow 0 < \cos θ < 1 \Rightarrow 0 < θ < \frac{π}{2} So, from equation (i), u = θ [Since, \sin^{- 1} (\sin θ) = θ, if θ \in [- \frac{π}{2}, \frac{π}{2}]] \Rightarrow u = \cos^{- 1} x$

Differentiating it with respect to x,

$\frac{d u}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} . . . (iii) From equation (ii), v = θ [Since, c o t^{- 1} (c o t θ) = θ, if θ \in (0, π)] \Rightarrow v = \cos^{- 1} x$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} . . . (iv) Dividing equation (iii) by (iv), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = (\frac{- 1}{\sqrt{1 - x^{2}}}) (\frac{\sqrt{1 - x^{2}}}{- 1}) ∴ \frac{d u}{d v} = 1$

Page No 10.113:

Question 19:

Differentiate $\sin^{- 1} (2 a x \sqrt{1 - a^{2} x^{2}})$ with respect to $\sqrt{1 - a^{2} x^{2}}, if - \frac{1}{\sqrt{2}} < a x < \frac{1}{\sqrt{2}}$

Answer:

$Let, u = \sin^{- 1} (2 a x \sqrt{1 - a^{2} x^{2}}) Put a x = \sin θ \Rightarrow θ = \sin^{- 1} (a x) ∴ u = \sin^{- 1} (2 \sin θ \sqrt{1 - \sin^{2} θ}) \Rightarrow u = \sin^{- 1} (2 \sin θ \cos θ) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) And Let, v = \sqrt{1 - a^{2} x^{2}} Differentiating it with respect to x, \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - a^{2} x^{2}}} \times \frac{d}{d x} (1 - a^{2} x^{2}) \Rightarrow \frac{d v}{d x} = (\frac{0 - 2 a^{2} x}{2 \sqrt{1 - a^{2} x^{2}}}) \Rightarrow \frac{d v}{d x} = \frac{- a^{2} x}{\sqrt{1 - a^{2} x^{2}}} . . . (ii) Here, - \frac{1}{\sqrt{2}} < a x < \frac{1}{\sqrt{2}} \Rightarrow - \frac{1}{\sqrt{2}} < \sin θ < \frac{1}{\sqrt{2}} \Rightarrow - \frac{π}{4} < θ < \frac{π}{4} So, from equation (i), u = 2 θ [Since, \sin^{- 1} (\sin θ) = θ, if θ \in [- \frac{π}{2}, \frac{π}{2}]] \Rightarrow u = 2 \sin^{- 1} x$

Differentiating it with respect to x,

$\frac{d u}{d x} = 2 \times \frac{1}{\sqrt{1 - {(a x)}^{2}}} \frac{d}{d x} (a x) \Rightarrow \frac{d u}{d x} = \frac{2}{1 - a^{2} x^{2}} (a) \Rightarrow \frac{d u}{d x} = \frac{2 a}{1 - a^{2} x^{2}} . . . (iii)$

$Dividing equation (iii) by (ii), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = (\frac{2 a}{\sqrt{1 - a^{2} x^{2}}}) (\frac{\sqrt{1 - a^{2} x^{2}}}{- a^{2} x}) ∴ \frac{d u}{d v} = - \frac{2}{a x}$

Page No 10.113:

Question 20:

Differentiate $\tan^{- 1} (\frac{1 - x}{1 + x})$ with respect to $\sqrt{1 - x^{2}}, if - 1 < x < 1$

Answer:

$Let, u = \tan^{- 1} (\frac{1 - x}{1 + x}) Put x = \tan θ \Rightarrow θ = \tan^{- 1} x \Rightarrow u = \tan^{- 1} (\frac{1 - \tan θ}{1 + \tan θ}) \Rightarrow u = \tan^{- 1} [\tan (\frac{π}{4} - θ)] . . . (i) Here, - 1 < x < 1 \Rightarrow - 1 < \tan θ < 1 \Rightarrow - \frac{π}{4} < θ < \frac{π}{4} \Rightarrow \frac{π}{4} > - θ > \frac{π}{4} \Rightarrow - \frac{π}{4} < - θ < \frac{π}{4} \Rightarrow 0 < \frac{π}{4} - θ < \frac{π}{2} So, from equation (i), u = \frac{π}{4} - θ [Since, \tan^{- 1} (\tan θ) = θ, if θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow u = \frac{π}{4} - \tan^{- 1} x$

Differentiating it with respect to x,

$\frac{d u}{d x} = 0 - (\frac{1}{1 + x^{2}}) \Rightarrow \frac{d u}{d x} = - \frac{1}{1 + x^{2}} . . . (ii) And let, v = \sqrt{1 - x^{2}}$

Differentiating it with respect to x,

$\frac{d v}{d x} = \frac{1}{2 \sqrt{1 - x^{2}}} \times \frac{d}{d x} (1 - x^{2}) \Rightarrow \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - x^{2}}} (- 2 x) \Rightarrow \frac{d v}{d x} = \frac{- x}{\sqrt{1 - x^{2}}} . . . (iii) Dividing equation (ii) by (iii), \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = - \frac{1}{1 + x^{2}} \times \frac{\sqrt{1 - x^{2}}}{- x} ∴ \frac{d u}{d v} = \frac{\sqrt{1 - x^{2}}}{x (1 + x^{2})}$

Page No 10.117:

Question 1:

If f (x) = log_x² (log x), the f' (x) at x = e is
(a) 0
(b) 1
(c) 1/e
(d) 1/2 e

Answer:

(d) 1/2 e

$We have, f (x) = \log_{x^{2}} (\log x) \Rightarrow f (x) = \frac{\log (\log x)}{\log x^{2}} \Rightarrow f (x) = \frac{\log (\log x)}{2 \log x} \Rightarrow f' (x) = \frac{1}{2} \times \frac{d}{d x} \{\frac{\log (\log x)}{\log x}\} \Rightarrow f' (x) = \frac{1}{2} \times \{\frac{\frac{1}{\log x} \times \frac{1}{x} \times \log x - \frac{\log (\log x)}{x}}{{(\log x)}^{2}}\} \Rightarrow f' (x) = \frac{1}{2} \times \{\frac{\frac{1}{x} - \frac{\log (\log x)}{x}}{{(\log x)}^{2}}\} \Rightarrow f' (e) = \frac{1}{2} \times \{\frac{\frac{1}{e} - \frac{\log (\log e)}{e}}{{(\log e)}^{2}}\} [Putting x = e] \Rightarrow f' (e) = \frac{1}{2} \times \{\frac{\frac{1}{e}}{1}\} \Rightarrow f' (e) = \frac{1}{2 e}$

Page No 10.117:

Question 2:

The differential coefficient of f (log x) w.r.t. x, where f (x) = log x is
(a) $\frac{x}{\log x}$

(b) $\frac{\log x}{x}$

(c) ${(x \log x)}^{- 1}$

(d) none of these

Answer:

(c) ${(x \log x)}^{- 1}$

We have,
$f (x) = \log x \Rightarrow f (\log x) = \log (\log x) \Rightarrow f' (\log x) = \frac{1}{\log x} \frac{d}{d x} (\log x) \Rightarrow f' (\log x) = \frac{1}{x \log x} \Rightarrow f' (\log x) = {(x \log x)}^{- 1}$

Page No 10.117:

Question 3:

The derivative of the function $\cot^{- 1} |{(\cos 2 x)}^{1 / 2}| at x = π / 6 is$
(a) (2/3)^1/2
(b) (1/3)^1/2
(c) 3^1/2
(d) 6^1/2

Answer:

(a) (2/3)^1/2
$We have, y = \cot^{- 1} (\sqrt{\cos 2 x})$

$\Rightarrow \frac{d y}{d x} = \frac{- 1}{1 + \cos 2 x} \frac{d}{d x} \sqrt{\cos 2 x} \Rightarrow \frac{d y}{d x} = \frac{- 1}{2 \cos^{2} x} \times \frac{1}{2 \sqrt{\cos 2 x}} \frac{d}{d x} (\cos 2 x) \Rightarrow \frac{d y}{d x} = \frac{- 1}{2 \cos^{2} x} \times \frac{1}{2 \sqrt{\cos 2 x}} \times - 2 \sin 2 x \Rightarrow \frac{d y}{d x} = \frac{\sin 2 x}{\cos^{2} x \times 2 \sqrt{\cos 2 x}} \Rightarrow \frac{d y}{d x} = \frac{2 \sin x \cos x}{\cos^{2} x \times 2 \sqrt{\cos 2 x}} \Rightarrow \frac{d y}{d x} = \frac{\tan x}{\sqrt{\cos 2 x}} So, at x = \frac{π}{6}, we get {(\frac{d y}{d x})}_{x = \frac{π}{6}} = \frac{\tan (\frac{π}{6})}{\sqrt{\cos 2 (\frac{π}{6})}} = \frac{(\frac{1}{\sqrt{3}})}{\sqrt{\frac{1}{2}}} = {(\frac{2}{3})}^{\frac{1}{2}}$

Page No 10.117:

Question 4:

Differential coefficient of sec $\sec (\tan^{- 1} x)$ is
(a) $\frac{x}{1 + x^{2}}$

(b) $x \sqrt{1 + x^{2}}$

(c) $\frac{1}{\sqrt{1 + x^{2}}}$

(d) $\frac{x}{\sqrt{1 + x^{2}}}$

Answer:

(d) $\frac{x}{\sqrt{1 + x^{2}}}$

$We have, y = \sec (\tan^{- 1} x) \Rightarrow \frac{d y}{d x} = \sec (\tan^{- 1} x) \tan (\tan^{- 1} x) \times \frac{d}{d x} (\tan^{- 1} x) \Rightarrow \frac{d y}{d x} = \sec (\tan^{- 1} x) \tan (\tan^{- 1} x) \times \frac{1}{\sqrt{1 + x^{2}}} \Rightarrow \frac{d y}{d x} = y (\frac{x}{\sqrt{1 + x^{2}}}) \Rightarrow \frac{d y}{d x} = (\frac{x}{\sqrt{1 + x^{2}}}) y$
This is the equation of differential equation which have coefficient $\frac{x}{\sqrt{1 + x^{2}}}$ .

Page No 10.117:

Question 5:

If $f (x) = \tan^{- 1} \sqrt{\frac{1 + \sin x}{1 - \sin x}}, 0 \leq x \leq π / 2, then f' (π / 6) is$
(a) − 1/4
(b) − 1/2
(c) 1/4
(d) 1/2

Answer:

(d) 1/2

$Let y = \tan^{- 1} \{\sqrt{\frac{1 + \sin x}{1 - \sin x}}\} \Rightarrow y = \tan^{- 1} \{\sqrt{\frac{1 - \cos (\frac{π}{2} + x)}{1 + \cos (\frac{π}{2} + x)}}\} \Rightarrow y = \tan^{- 1} \{\sqrt{\frac{2 \sin^{2} (\frac{π}{4} + \frac{x}{2})}{2 \cos^{2} (\frac{π}{4} + \frac{x}{2})}}\} \Rightarrow y = \tan^{- 1} \{\tan (\frac{π}{4} + \frac{x}{2})\} = \frac{π}{4} + \frac{x}{2} ∴ \frac{d y}{d x} = \frac{1}{2}$

Page No 10.117:

Question 6:

If $y = {(1 + \frac{1}{x})}^{x}, then \frac{d y}{d x} =$

(a) ${(1 + \frac{1}{x})}^{x} (1 + \frac{1}{x}) - \frac{1}{x + 1}$

(b) ${(1 + \frac{1}{x})}^{x} \log (1 + \frac{1}{x})$

(c) ${(x + \frac{1}{x})}^{x} \{\log (x + 1) - \frac{x}{x + 1}\}$

(d) ${(x + \frac{1}{x})}^{x} \{\log (1 + \frac{1}{x}) + \frac{1}{x + 1}\}$

Answer:

(a) ${(1 + \frac{1}{x})}^{x} (1 + \frac{1}{x}) - \frac{1}{x + 1}$

$Let y = {(1 + \frac{1}{x})}^{x} Taking \log on both sides, \log y = x \log (1 + \frac{1}{x}) \Rightarrow \frac{1}{y} \frac{d y}{d x} = x \frac{d}{d x} \log (1 + \frac{1}{x}) + \log (1 + \frac{1}{x}) \frac{d}{d x} (x) \Rightarrow \frac{1}{y} \frac{d y}{d x} = x (\frac{1}{1 + \frac{1}{x}}) \frac{d}{d x} (1 + \frac{1}{x}) + \log (1 + \frac{1}{x}) \Rightarrow \frac{1}{y} \frac{d y}{d x} = x \times \frac{x}{x + 1} (- \frac{1}{x^{2}}) + \log (1 + \frac{1}{x}) \Rightarrow \frac{1}{y} \frac{d y}{d x} = \frac{x^{2}}{x + 1} \times \frac{- 1}{x^{2}} + \log (1 + \frac{1}{x}) \Rightarrow \frac{1}{y} \frac{d y}{d x} = \frac{- 1}{x + 1} + \log (1 + \frac{1}{x}) \Rightarrow \frac{d y}{d x} = y [\frac{- 1}{x + 1} + \log (1 + \frac{1}{x})] \Rightarrow \frac{d y}{d x} = {(1 + \frac{1}{x})}^{x} [\log (1 + \frac{1}{x}) - \frac{1}{x + 1}]$

Page No 10.118:

Question 7:

If $x^{y} = e^{x - y}, then \frac{d y}{d x}$ is
(a) $\frac{1 + x}{1 + \log x}$

(b) $\frac{1 - \log x}{1 + \log x}$

(c) not defined

(d) $\frac{\log x}{{(1 + \log x)}^{2}}$

Answer:

(d) $\frac{\log x}{{(1 + \log x)}^{2}}$

$We have, x^{y} = e^{x - y} Taking \log on both sides we get, \Rightarrow y \log x = (x - y) \log_{e} e \Rightarrow y \log x = x - y \Rightarrow y \log x + y = x \Rightarrow y (1 + \log x) = x \Rightarrow y = \frac{x}{1 + \log x}$
$\Rightarrow \frac{d y}{d x} = \frac{(1 + \log x) \times 1 - x \times (0 + \frac{1}{x})}{{(1 + \log x)}^{2}} \Rightarrow \frac{d y}{d x} = \frac{1 + \log x - 1}{{(1 + \log x)}^{2}} \Rightarrow \frac{d y}{d x} = \frac{\log x}{{(1 + \log x)}^{2}}$

Page No 10.118:

Question 8:

Given $f (x) = 4 x^{8}, then$

(a) $f' (\frac{1}{2}) = f' (- \frac{1}{2})$

(b) $f (\frac{1}{2}) = - f' (- \frac{1}{2})$

(c) $f (- \frac{1}{2}) = f (- \frac{1}{2})$

(d) $f (\frac{1}{2}) = f' (- \frac{1}{2})$

Answer:

$(c) f (\frac{- 1}{2}) = f (\frac{- 1}{2}) We have, f (x) = 4 x^{8} \Rightarrow f' (x) = 32 x^{7} Now, f (\frac{1}{2}) = 4 {(\frac{1}{2})}^{8} = 4 (\frac{1}{256}) = \frac{1}{64} f (- \frac{1}{2}) = 4 {(- \frac{1}{2})}^{8} = 4 (\frac{1}{256}) = \frac{1}{64} f' (\frac{1}{2}) = 32 {(\frac{1}{2})}^{7} = 32 (\frac{1}{128}) = \frac{1}{4} f' (- \frac{1}{2}) = 32 {(- \frac{1}{2})}^{7} = - 32 (\frac{1}{128}) = - \frac{1}{4}$

Page No 10.118:

Question 9:

If $x = a \cos^{3} θ, y = a \sin^{3} θ, then \sqrt{1 + {(\frac{d y}{d x})}^{2}} =$
(a) $\tan^{2} θ$
(b) $\sec^{2} θ$
(c) $\sec θ$
(d) $|\sec θ|$

Answer:

(d) $|\sec θ|$

$We have, x = a \cos^{3} θ \Rightarrow \frac{d x}{d θ} = a \frac{d}{d θ} (\cos^{3} θ) \Rightarrow \frac{d x}{d θ} = 3 a \cos^{2} θ \frac{d}{d θ} (\cos θ) \Rightarrow \frac{d x}{d θ} = - 3 a \cos^{2} θ \sin θ . . . (1) and, y = a \sin^{3} θ \Rightarrow \frac{d y}{d θ} = a \frac{d}{d θ} (\sin^{3} θ) \Rightarrow \frac{d y}{d θ} = 3 a \sin^{2} θ \frac{d}{d θ} (\sin θ) \Rightarrow \frac{d y}{d θ} = 3 a \sin^{2} θ \cos θ . . . (2) Dividing (2) by (1), we get, \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{3 a \sin^{2} θ \cos θ}{- 3 a \cos^{2} θ \sin θ} \Rightarrow \frac{d y}{d x} = \frac{\sin θ}{- \cos θ} \Rightarrow \frac{d y}{d x} = - \tan θ Now, \sqrt{1 + {(\frac{d y}{d x})}^{2}} = \sqrt{1 + \tan^{2} θ} = \sqrt{\sec^{2} θ} = |\sec θ|$

Page No 10.118:

Question 10:

If $y = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), then \frac{d y}{d x} =$

(a) $- \frac{2}{1 + x^{2}}$

(b) $\frac{2}{1 + x^{2}}$

(c) $\frac{1}{2 - x^{2}}$

(d) $\frac{2}{2 - x^{2}}$

Answer:

(a) $- \frac{2}{1 + x^{2}}$

$Let y = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) Differentiating with respect to x using chain rule, we get, \frac{d y}{d x} = \frac{1}{\sqrt{1 - {(\frac{1 - x^{2}}{1 + x^{2}})}^{2}}} \frac{d}{d x} (\frac{1 - x^{2}}{1 + x^{2}}) \Rightarrow \frac{d y}{d x} = \frac{(1 + x^{2})}{\sqrt{{(1 + x^{2})}^{2} - {(1 - x^{2})}^{2}}} [\frac{(1 + x^{2}) \frac{d}{d x} (1 - x^{2}) - (1 - x^{2}) \frac{d}{d x} (1 + x^{2})}{{(1 + x^{2})}^{2}}] [using quotient rule] \Rightarrow \frac{d y}{d x} = \frac{(1 + x^{2})}{\sqrt{{(1 + x^{2})}^{2} - {(1 - x^{2})}^{2}}} [\frac{(1 + x^{2}) (- 2 x) - (1 - x^{2}) (2 x)}{{(1 + x^{2})}^{2}}] \Rightarrow \frac{d y}{d x} = \frac{(1 + x^{2})}{2 x} [\frac{- 2 x - 2 x^{3} - 2 x + 2 x^{3}}{{(1 + x^{2})}^{2}}] \Rightarrow \frac{d y}{d x} = \frac{- 4 x}{2 x (1 + x^{2})} \Rightarrow \frac{d y}{d x} = \frac{- 2}{1 + x^{2}}$

Page No 10.118:

Question 11:

The derivative of $\sec^{- 1} (\frac{1}{2 x^{2} + 1}) w . r . t . \sqrt{1 + 3 x} at x = - 1 / 3$

(a) does not exist
(b) 0
(c) 1/2
(d) 1/3

Answer:

(a) does not exist

$We know that \sec^{- 1} α is not defined for α \in (- 1, 1) Here for x = \frac{- 1}{3}, \frac{1}{2 x^{2} + 1} = \frac{9}{11} \in (- 1, 1) ∴ \sec^{- 1} (\frac{1}{2 x^{2} + 1}) is not defined at x = \frac{- 1}{3} ∴ Derivative of \sec^{- 1} (\frac{1}{2 x^{2} + 1}) does not exist at x = \frac{- 1}{3}$

Page No 10.118:

Question 12:

For the curve $\sqrt{x} + \sqrt{y} = 1, \frac{d y}{d x} at (1 / 4, 1 / 4) is$
(a) 1/2
(b) 1
(c) −1
(d) 2

Answer:

(c) −1

$We have, \sqrt{x} + \sqrt{y} = 1 Differentiating with respect to x, we get, \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0 \Rightarrow \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = - \frac{1}{2 \sqrt{x}} \Rightarrow \frac{d y}{d x} = - \frac{1}{2 \sqrt{x}} \times \frac{2 \sqrt{y}}{1} \Rightarrow \frac{d y}{d x} = - \frac{\sqrt{y}}{\sqrt{x}} Now, {[\frac{d y}{d x}]}_{(\frac{1}{4}, \frac{1}{4})} = - \frac{\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}} = - 1$

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Question 13:

If $\sin (x + y) = \log (x + y), then \frac{d y}{d x} =$
(a) 2
(b) − 2
(c) 1
(d) − 1]

Answer:

(d) − 1

$We have, \sin (x + y) = \log (x + y) \Rightarrow \cos (x + y) (1 + \frac{d y}{d x}) = \frac{1}{(x + y)} (1 + \frac{d y}{d x}) \Rightarrow \cos (x + y) + \cos (x + y) \frac{d y}{d x} = \frac{1}{(x + y)} + \frac{1}{(x + y)} \frac{d y}{d x} \Rightarrow \cos (x + y) \frac{d y}{d x} - \frac{1}{(x + y)} \frac{d y}{d x} = \frac{1}{(x + y)} - \cos (x + y) \Rightarrow \{\cos (x + y) - \frac{1}{(x + y)}\} \frac{d y}{d x} = \frac{1}{(x + y)} - \cos (x + y) \Rightarrow - \{\frac{1}{(x + y)} - \cos (x + y)\} \frac{d y}{d x} = \frac{1}{(x + y)} - \cos (x + y) \Rightarrow \frac{d y}{d x} = - 1$

Page No 10.118:

Question 14:

Let $\cup = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) and V = \tan^{- 1} (\frac{2 x}{1 - x^{2}}), then \frac{d \cup}{d V} =$
(a) 1/2
(b) x
(c) $\frac{1 - x^{2}}{x^{2} - 4}$
(d) 1

Answer:

(d) 1
$We have, u = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) and v = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) \Rightarrow \frac{d u}{d x} = \frac{2}{1 + x^{2}} and \frac{d v}{d x} = \frac{2}{1 + x^{2}}$

$∴ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{2}{1 + x^{2}} \times \frac{1 + x^{2}}{2} = 1$

Page No 10.118:

Question 15:

$\frac{d}{d x} \{\tan^{- 1} (\frac{\cos x}{1 + \sin x})\} equals$

(a) 1/2
(b) − 1/2
(c) 1
(d) − 1

Answer:

(b) − 1/2

$Let u = \tan^{- 1} (\frac{\cos x}{1 + \sin x}) \Rightarrow u = \tan^{- 1} (\frac{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}}) \Rightarrow u = \tan^{- 1} \frac{(\cos \frac{x}{2} - \sin \frac{x}{2}) (\cos \frac{x}{2} + \sin \frac{x}{2})}{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{2}} \Rightarrow u = \tan^{- 1} (\frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}}) \Rightarrow u = \tan^{- 1} [\frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}}] \Rightarrow u = \tan^{- 1} [\frac{\tan \frac{π}{4} - \tan \frac{x}{2}}{1 + \tan \frac{π}{4} \times \tan \frac{x}{2}}] \Rightarrow u = \tan^{- 1} [\tan (\frac{π}{4} - \frac{x}{2})] \Rightarrow u = \frac{π}{4} - \frac{x}{2}$

$\Rightarrow \frac{d u}{d x} = 0 - (\frac{1}{2}) \Rightarrow \frac{d u}{d x} = - \frac{1}{2}$

Page No 10.119:

Question 16:

$\frac{d}{d x} [\log \{e^{x} {(\frac{x - 2}{x + 2})}^{3 / 4}\}]$ equals

(a) $\frac{x^{2} - 1}{x^{2} - 4}$

(b) 1

(c) $\frac{x^{2} + 1}{x^{2} - 4}$

(d) $e^{x} \frac{x^{2} - 1}{x^{2} - 4}$

Answer:

(a) $\frac{x^{2} - 1}{x^{2} - 4}$

$Let y = \frac{d}{d x} [\log \{e^{x} {(\frac{x - 2}{x + 2})}^{\frac{3}{4}}\}] \Rightarrow y = \frac{d}{d x} [x \log e + \frac{3}{4} \log (\frac{x - 2}{x + 2})] \Rightarrow y = \frac{d}{d x} [x + \frac{3}{4} \log (\frac{x - 2}{x + 2})] \Rightarrow \frac{d y}{d x} = 1 + \frac{3}{4 (\frac{x - 2}{x + 2})} \times \frac{(x + 2) \times 1 - (x - 2) \times 1}{{(x + 2)}^{2}} \Rightarrow \frac{d y}{d x} = 1 + \frac{3 (x + 2)}{4 (x - 2)} \times \frac{x + 2 - x + 2}{{(x + 2)}^{2}} \Rightarrow \frac{d y}{d x} = 1 + \frac{3 (x + 2)}{4 (x - 2)} \times \frac{4}{(x + 2)} \Rightarrow \frac{d y}{d x} = 1 + \frac{3}{(x^{2} - 4)} \Rightarrow \frac{d y}{d x} = \frac{x^{2} - 4 + 3}{x^{2} - 4} \Rightarrow \frac{d y}{d x} = \frac{x^{2} - 1}{x^{2} - 4}$

Page No 10.119:

Question 17:

If $y = \sqrt{\sin x + y}, then \frac{d y}{d x} =$

(a) $\frac{\sin x}{2 y - 1}$

(b) $\frac{\sin x}{1 - 2 y}$

(c) $\frac{\cos x}{1 - 2 y}$

(d) $\frac{\cos x}{2 y - 1}$

Answer:

(d) $\frac{\cos x}{2 y - 1}$

$We have, y = \sqrt{\sin x + y} Squaring both sides, we get, y^{2} = \sin x + y \Rightarrow y^{2} - y = \sin x \Rightarrow 2 y \frac{d y}{d x} - \frac{d y}{d x} = \cos x \Rightarrow \frac{d y}{d x} (2 y - 1) = \cos x \Rightarrow \frac{d y}{d x} = \frac{\cos x}{2 y - 1}$

Page No 10.119:

Question 18:

If $3 \sin (x y) + 4 \cos (x y) = 5, then \frac{d y}{d x} =$

(a) $- \frac{y}{x}$

(b) $\frac{3 \sin (x y) + 4 \cos (x y)}{3 \cos (x y) - 4 \sin (x y)}$

(c) $\frac{3 \cos (x y) + 4 \sin (x y)}{4 \cos (x y) - 3 \sin (x y)}$

(d) none of these

Answer:

(a) $- \frac{y}{x}$

$We have, 3 \sin (x y) + 4 \cos (x y) = 5 \Rightarrow 3 \cos (x y) [x \frac{d y}{d x} + y] - 4 \sin (x y) [x \frac{d y}{d x} + y] = 0 \Rightarrow [x \frac{d y}{d x} + y] [3 \cos (x y) - 4 \sin (x y)] = 0 \Rightarrow x \frac{d y}{d x} + y = 0 \Rightarrow x \frac{d y}{d x} = - y ∴ \frac{d y}{d x} = - \frac{y}{x}$

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Question 19:

If $\sin y = x \sin (a + y), then \frac{d y}{d x} is$

(a) $\frac{\sin a}{\sin a \sin^{2} (a + y)}$

(b) $\frac{\sin^{2} (a + y)}{\sin a}$

(c) $\sin a \sin^{2} (a + y)$

(d) $\frac{\sin^{2} (a - y)}{\sin a}$

Answer:

(b) $\frac{\sin^{2} (a + y)}{\sin a}$

$We have, \sin y = x \sin (a + y)$

$\Rightarrow \frac{d}{d x} (\sin y) = \frac{d}{d x} [x \sin (a + y)] \Rightarrow \cos y \frac{d y}{d x} = \sin (a + y) \frac{d}{d x} (x) + x \frac{d}{d x} \{\sin (a + y)\} \Rightarrow \cos y \frac{d y}{d x} = \sin (a + y) \times 1 + x \cos (a + y) \frac{d y}{d x} \Rightarrow \cos y \frac{d y}{d x} = \sin (a + y) + x \cos (a + y) \frac{d y}{d x} \Rightarrow \cos y \frac{d y}{d x} - x \cos (a + y) \frac{d y}{d x} = \sin (a + y) \Rightarrow \{\cos y - x \cos (a + y)\} \frac{d y}{d x} = \sin (a + y) \Rightarrow \{\cos y - \frac{\sin y}{\sin (a + y)} \times \cos (a + y)\} \frac{d y}{d x} = \sin (a + y) [\begin{matrix} ∵ \sin y = 2 \sin x \cos x \\ ∴ x = \frac{\sin y}{\sin (a + y)} \end{matrix}] \Rightarrow \{\frac{\sin (a + y) \cos y - \sin y \cos (a + y)}{\sin (a + y)}\} \frac{d y}{d x} = \sin (a + y) \Rightarrow \frac{\sin (a + y - y)}{\sin (a + y)} \times \frac{d y}{d x} = \sin (a + y) \Rightarrow \frac{d y}{d x} = \frac{\sin^{2} (a + y)}{\sin a}$

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Question 20:

The derivative of $\cos^{- 1} (2 x^{2} - 1)$ with respect to $\cos^{- 1} x$ is
(a) 2

(b) $\frac{1}{2 \sqrt{1 - x^{2}}}$

(c) $2 / x$

(d) $1 - x^{2}$

Answer:

(a) 2

$Let u = \cos^{- 1} (2 x^{2} - 1) Put x = \cos θ \Rightarrow θ = \cos^{- 1} x \frac{d θ}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} Now, u = \cos^{- 1} (\cos 2 θ) \Rightarrow u = 2 θ$

$\Rightarrow \frac{d u}{d x} = 2 \frac{d θ}{d x} \Rightarrow \frac{d u}{d x} = \frac{- 2}{\sqrt{1 - x^{2}}} . . . (i) and, v = \cos^{- 1} x \Rightarrow v = \cos^{- 1} (\cos θ) \Rightarrow v = θ$

$\frac{d v}{d x} = \frac{d θ}{d x} \Rightarrow \frac{d v}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} . . . (i i) Dividing (i) by (ii), we get, \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{- 2}{\sqrt{1 - x^{2}}} \times \frac{\sqrt{1 - x^{2}}}{- 1} \Rightarrow \frac{d u}{d v} = 2$

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Question 21:

If $f (x) = \sqrt{x^{2} + 6 x + 9}, then f' (x)$ is equal to
(a) $1 for x < - 3$
(b) $- 1 for x < - 3$
(c) $1 for all x \in R$
(d) none of these

Answer:

$(b) - 1 for x < - 3 We have, f (x) = \sqrt{x^{2} + 6 x + 9} = \sqrt{{(x + 3)}^{2}} = |x + 3| f (x) = \{\begin{cases} x + 3, x \geq - 3 \\ - x - 3, x < - 3 \end{cases} \Rightarrow f' (x) = \{\begin{cases} 1, x \geq - 3 \\ - 1, x < - 3 \end{cases} ∴ f' (x) = - 1 for x < - 3$

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Question 22:

If $f (x) = |x^{2} - 9 x + 20|$ , then f' (x) is equal to
(a) $- 2 x + 9 for all x \in R$
(b) $2 x - 9 if 4 < x < 5$
(c) $- 2 x + 9, if 4 < x < 5$
(d) none of these

Answer:

$(c) - 2 x + 9 for 4 < x < 5 We have, f (x) = |x^{2} - 9 x + 20| f (x) = \{\begin{matrix} x^{2} - 9 x + 20, - \infty < x \leq 4 \\ - (x^{2} - 9 x + 20), 4 < x < 5 \\ x^{2} - 9 x + 20, 5 \leq x < \infty \end{matrix}\} \Rightarrow f (x) = \{\begin{matrix} 2 x - 9, - \infty < x \leq 4 \\ - 2 x + 9, 4 < x < 5 \\ 2 x - 9, 5 \leq x < \infty \end{matrix}\} ∴ f' (x) = - 2 x + 9 for 4 < x < 5$

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Question 23:

If $f (x) = \sqrt{x^{2} - 10 x + 25}$ , then the derivative of f (x) in the interval [0, 7] is
(a) 1
(b) −1
(c) 0
(d) none of these

Answer:

(d) none of these

$We have, f (x) = \sqrt{x^{2} - 10 x + 25} = \sqrt{{(x - 5)}^{2}} = |x - 5| = \{\begin{cases} x - 5 for x > 5 \\ - (x - 5) for x < 5 \end{cases} LHD = \lim_{x \to 5^{-}} \frac{f (x) - f (a)}{x - a} = \lim_{x \to 5^{-}} \frac{\sqrt{x^{2} - 10 x + 25} - \sqrt{5^{2} - 10 (5) + 25}}{x - 5} = \lim_{x \to 5^{-}} \frac{|x - 5|}{x - 5} = \lim_{x \to 5^{-}} \frac{- (x - 5)}{x - 5} = - 1 RHD = \lim_{x \to 5^{+}} \frac{f (x) - f (a)}{x - a} = \lim_{x \to 5^{+}} \frac{\sqrt{x^{2} - 10 x + 25} - \sqrt{5^{2} - 10 (5) + 25}}{x - 5} = \lim_{x \to 5^{+}} \frac{|x - 5|}{x - 5} = \lim_{x \to 5^{+}} \frac{x - 5}{x - 5} = 1 Here, LHD \neq RHD Thus, the functionis not differentiable at x = 5$

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Question 24:

If $f (x) = |x - 3| and g (x) = f o f (x)$ , then for x > 10, g ' (x) is equal to
(a) 1
(b) −1
(c) 0
(d) none of these

Answer:

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Question 25:

If $f (x) = {(\frac{x^{l}}{x^{m}})}^{l + m} {(\frac{x^{m}}{x^{n}})}^{m + n} {(\frac{x^{n}}{x^{l}})}^{n + 1}$ , the f' (x) is equal to
(a) 1
(b) 0
(c) $x^{l + m + n}$
(d) none of these

Answer:

(b) 0
We have, $f (x) = {(\frac{x^{l}}{x^{m}})}^{l + m} {(\frac{x^{m}}{x^{n}})}^{m + n} {(\frac{x^{n}}{x^{l}})}^{n + 1}$

$\Rightarrow f (x) = x^{(l - m) (l + m)} \times x^{(m - n) (m + n)} \times x^{(n - l) (n + l)} \Rightarrow f (x) = x^{l^{2} - m^{2}} \times x^{m^{2} - n^{2}} \times x^{n^{2} - l^{2}} \Rightarrow f (x) = x^{(l^{2} - m^{2} + m^{2} - n^{2} + n^{2} - l^{2})} \Rightarrow f (x) = x^{0} \Rightarrow f (x) = 1 \Rightarrow f' (x) = 0$

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Question 26:

If $y = \frac{1}{1 + x^{a - b} +^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} + \frac{1}{1 + x^{b - a} + x^{c - a}}$ , then $\frac{d y}{d x}$ is equal to
(a) 1
(b) ${(a + b + c)}^{x^{a + b + c - 1}}$
(c) 0
(d) none of these

Answer:

(c) 0

$We have, y = \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} + \frac{1}{1 + x^{b - a} + x^{c - a}} = \frac{1}{1 + \frac{x^{a}}{x^{b}} + \frac{x^{c}}{x^{b}}} + \frac{1}{1 + \frac{x^{b}}{x^{c}} + \frac{x^{a}}{x^{c}}} + \frac{1}{1 + \frac{x^{b}}{x^{a}} + \frac{x^{c}}{x^{a}}} = \frac{x^{b}}{x^{b} + x^{a} + x^{c}} + \frac{x^{c}}{x^{c} + x^{b} + x^{a}} + \frac{x^{a}}{x^{a} + x^{b} + x^{c}} = \frac{x^{b}}{x^{a} + x^{b} + x^{c}} + \frac{x^{c}}{x^{a} + x^{b} + x^{c}} + \frac{x^{a}}{x^{a} + x^{b} + x^{c}} = \frac{x^{b} + x^{c} + x^{a}}{x^{a} + x^{b} + x^{c}} = \frac{x^{a} + x^{b} + x^{c}}{x^{a} + x^{b} + x^{c}} = 1 ∴ \frac{d y}{d x} = \frac{d}{d x} (1) = 0$

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Question 27:

If $\sqrt{1 - x^{6}} + \sqrt{1 - y^{6}} = a^{3} (x^{3} - y^{3})$ , then $\frac{d y}{d x}$ is equal to
(a) $\frac{x^{2}}{y^{2}} \sqrt{\frac{1 - y^{6}}{1 - x^{6}}}$

(b) $\frac{y^{2}}{x^{2}} \sqrt{\frac{1 - y^{6}}{1 + x^{6}}}$

(c) $\frac{x^{2}}{y^{2}} \sqrt{\frac{1 - x^{6}}{1 - y^{6}}}$

(d) none of these

Answer:

(a) $\frac{x^{2}}{y^{2}} \sqrt{\frac{1 - y^{6}}{1 - x^{6}}}$

$We have, \sqrt{1 - x^{6}} + \sqrt{1 - y^{6}} = a (x^{3} - y^{3}) Putting x^{3} = \sin A and y^{3} = \sin B \Rightarrow \sqrt{1 - \sin^{2} A} + \sqrt{1 - \sin^{2} B} = a (\sin A - \sin B) \Rightarrow \cos A + \cos B = a (\sin A - \sin B) \Rightarrow 2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2}) = 2 a \sin (\frac{A - B}{2}) \cos (\frac{A + B}{2}) \Rightarrow \cot (\frac{A - B}{2}) = a \Rightarrow \frac{A - B}{2} = \cot^{- 1} (a) \Rightarrow A - B = 2 \cot^{- 1} (a) \Rightarrow \sin^{- 1} x^{3} - \sin^{- 1} y^{3} = 2 \cot^{- 1} (a)$

$\Rightarrow \frac{1}{\sqrt{1 - x^{6}}} \times \frac{d}{d x} (x^{3}) - \frac{1}{\sqrt{1 - y^{6}}} \times \frac{d}{d x} (y^{3}) = 0 \Rightarrow \frac{1}{\sqrt{1 - x^{6}}} \times 3 x^{2} - \frac{1}{\sqrt{1 - y^{6}}} \times 3 y^{2} \times \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{x^{2}}{y^{2}} \sqrt{\frac{1 - y^{6}}{1 - x^{6}}}$

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Question 28:

If $y = \log \sqrt{\tan x}$ , then the value of $\frac{d y}{d x} at x = \frac{π}{4}$ is given by
(a) ∞
(b) 1
(c) 0
(d) $\frac{1}{2}$

Answer:

(b) 1

$We have, y = \log \sqrt{\tan x} \Rightarrow \frac{d y}{d x} = \frac{1}{\sqrt{\tan x}} \times \frac{d}{d x} (\sqrt{\tan x}) \Rightarrow \frac{d y}{d x} = \frac{1}{\sqrt{\tan x}} \times \frac{1}{2 \sqrt{\tan x}} \times \frac{d}{d x} (\tan x) \Rightarrow \frac{d y}{d x} = \frac{\sec^{2} x}{2 \tan x} Now, {(\frac{d y}{d x})}_{x = \frac{π}{4}} = \frac{{[\sec (\frac{π}{4})]}^{2}}{2 \tan (\frac{π}{4})} = \frac{2}{2 \times 1} = 1$

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Question 29:

If $\sin^{- 1} (\frac{x^{2} - y^{2}}{x^{2} + y^{2}}) = \log a then \frac{d y}{d x}$ is equal to

(a) $\frac{x^{2} - y^{2}}{x^{2} + y^{2}}$

(b) $\frac{y}{x}$

(c) $\frac{x}{y}$

(d) none of these

Answer:

(b) $\frac{y}{x}$
$We have, \sin^{- 1} (\frac{x^{2} - y^{2}}{x^{2} + y^{2}}) = \log a \Rightarrow \frac{x^{2} - y^{2}}{x^{2} + y^{2}} = \sin \log a$

$\Rightarrow \frac{(x^{2} + y^{2}) (2 x - 2 y \frac{d y}{d x}) - (x^{2} - y^{2}) (2 x + 2 y \frac{d y}{d x})}{{(x^{2} + y^{2})}^{2}} = 0 \Rightarrow \frac{2 x^{3} - 2 x^{2} y \frac{d y}{d x} + 2 x y^{2} - 2 y^{3} \frac{d y}{d x} - 2 x^{3} - 2 x^{2} y \frac{d y}{d x} + 2 x y^{2} + 2 y^{3} \frac{d y}{d x}}{{(x^{2} + y^{2})}^{2}} = 0 \Rightarrow - 4 x^{2} y \frac{d y}{d x} + 4 x y^{2} = 0 \Rightarrow - 4 x^{2} y \frac{d y}{d x} = - 4 x y^{2} \Rightarrow \frac{d y}{d x} = \frac{4 x y^{2}}{4 x^{2} y} ∴ \frac{d y}{d x} = \frac{y}{x}$

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Question 30:

If $\sin y = x \cos (a + y), then \frac{d y}{d x}$ is equal to
(a) $\frac{\cos^{2} (a + y)}{\cos a}$

(b) $\frac{\cos a}{\cos^{2} (a + y)}$

(c) $\frac{\sin^{2} y}{\cos a}$

(d) none of these

Answer:

(a) $\frac{\cos^{2} (a + y)}{\cos a}$

We have, $\sin y = x \cos (a + y)$

$\Rightarrow \frac{d}{d x} (\sin y) = \frac{d}{d x} [x \cos (a + y)] \Rightarrow \cos y \frac{d y}{d x} = 1 \times \cos (a + y) - x \sin (a + y) \frac{d}{d x} (a + y) \Rightarrow \cos y \frac{d y}{d x} = \cos (a + y) - x \sin (a + y) \frac{d y}{d x} \Rightarrow \cos y \frac{d y}{d x} + x \sin (a + y) \frac{d y}{d x} = \cos (a + y) \Rightarrow [\cos y + x \sin (a + y)] \frac{d y}{d x} = \cos (a + y) \Rightarrow [\cos y + \frac{\sin y}{\cos (a + y)} \times \sin (a + y)] \frac{d y}{d x} = \cos (a + y) [\begin{matrix} ∵ \sin y = x \cos (a + y) \\ ∵ x = \frac{\sin y}{\cos (a + y)} \end{matrix}] \Rightarrow [\frac{\cos (a + y) \cos y + \sin y \sin (a + y)}{\cos (a + y)}] \frac{d y}{d x} = \cos (a + y) \Rightarrow \frac{\cos (a + y - y)}{\cos (a + y)} \times \frac{d y}{d x} = \cos (a + y) \Rightarrow \frac{d y}{d x} = \frac{\cos^{2} (a + y)}{\cos a}$

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Question 31:

If $y = \log (\frac{1 - x^{2}}{1 + x^{2}}), then \frac{d y}{d x} =$

(a) $\frac{4 x^{3}}{1 - x^{4}}$

(b) $- \frac{4 x}{1 - x^{4}}$

(c) $\frac{1}{4 - x^{4}}$

(d) $- \frac{4 x^{3}}{1 - x^{4}}$

Answer:

(b) $- \frac{4 x}{1 - x^{4}}$
$We have, y = \log (\frac{1 - x^{2}}{1 + x^{2}}) \Rightarrow \frac{d y}{d x} = \frac{1}{\frac{1 - x^{2}}{1 + x^{2}}} \frac{d}{d x} (\frac{1 - x^{2}}{1 + x^{2}}) \Rightarrow \frac{d y}{d x} = \frac{1 + x^{2}}{1 - x^{2}} [\frac{(1 + x^{2}) (- 2 x) - (1 - x^{2}) (2 x)}{{(1 + x^{2})}^{2}}] \Rightarrow \frac{d y}{d x} = \frac{1}{1 - x^{2}} [\frac{- 2 x - 2 x^{3} - 2 x + 2 x^{3}}{(1 + x^{2})}] \Rightarrow \frac{d y}{d x} = \frac{- 4 x}{1 - x^{4}}$

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Question 32:

If $y = \tan^{- 1} (\frac{\sin x + \cos x}{\cos x - \sin x}), then \frac{d y}{d x}$ is equal to
(a) $\frac{1}{2}$
(b) 0
(c) 1
(d) none of these

Answer:

(c) 1

$We have, y = \tan^{- 1} (\frac{\sin x + \cos x}{\cos x - \sin x}) \Rightarrow \frac{d y}{d x} = \frac{1}{1 + {(\frac{\sin x + \cos x}{\cos x - \sin x})}^{2}} \frac{d}{d x} (\frac{\sin x + \cos x}{\cos x - \sin x}) \Rightarrow \frac{d y}{d x} = \frac{{(\cos x - \sin x)}^{2}}{{(\cos x - \sin x)}^{2} + {(\sin x + \cos x)}^{2}} [\frac{(\cos x - \sin x) \frac{d}{d x} (\sin x + \cos x) - (\sin x + \cos x) \frac{d}{d x} (\cos x - \sin x)}{{(\cos x - \sin x)}^{2}}] \Rightarrow \frac{d y}{d x} = \frac{{(\cos x - \sin x)}^{2}}{{(\cos x - \sin x)}^{2} + {(\sin x + \cos x)}^{2}} [\frac{(\cos x - \sin x) (\cos x - \sin x) - (\sin x + \cos x) (- \sin x - \cos x)}{{(\cos x - \sin x)}^{2}}] \Rightarrow \frac{d y}{d x} = \frac{{(\cos x - \sin x)}^{2}}{{(\cos x - \sin x)}^{2} + {(\sin x + \cos x)}^{2}} [\frac{(\cos x - \sin x) (\cos x - \sin x) + (\sin x + \cos x) (\sin x + \cos x)}{{(\cos x - \sin x)}^{2}}] \Rightarrow \frac{d y}{d x} = \frac{{(\cos x - \sin x)}^{2}}{{(\cos x - \sin x)}^{2} + {(\sin x + \cos x)}^{2}} \times \frac{{(\cos x - \sin x)}^{2} + {(\sin x + \cos x)}^{2}}{{(\cos x - \sin x)}^{2}} \Rightarrow \frac{d y}{d x} = 1$

Page No 10.120:

Question 1:

If y = x $|x|, then {(\frac{d y}{d x})}_{x = - 1} =______________________.$

Answer:

We know

$|x| = \{\begin{cases} x, & x \geq 0 \\ - x, & x < 0 \end{cases}$

$∴ y = x |x| = \{\begin{cases} x^{2}, & x \geq 0 \\ - x^{2}, & x < 0 \end{cases}$

$\Rightarrow \frac{d y}{d x} = \{\begin{cases} 2 x, & x \geq 0 \\ - 2 x, & x < 0 \end{cases}$

For x < 0,

$\frac{d y}{d x} = - 2 x$

$∴ {(\frac{d y}{d x})}_{x = - 1} = - 2 \times (- 1) = 2$

If y = x $|x|, then {(\frac{d y}{d x})}_{x = - 1} = 2 .$

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Question 2:

If y = 2x + $|x|, then {(\frac{d y}{d x})}_{x = - 1} =_________________and {(\frac{d y}{d x})}_{x = 1} =____________________.$

Answer:

We know

$|x| = \{\begin{cases} x, & x \geq 0 \\ - x, & x < 0 \end{cases}$

$∴ y = 2 x + |x| = \{\begin{cases} 2 x + x, & x \geq 0 \\ 2 x - x, & x < 0 \end{cases}$

$\Rightarrow y = 2 x + |x| = \{\begin{cases} 3 x, & x \geq 0 \\ x, & x < 0 \end{cases}$

For x ≥ 0,

y = 3x

$\Rightarrow \frac{d y}{d x} = 3$

$∴ {(\frac{d y}{d x})}_{x = 1} = 3$

For x < 0,

y = x

$\Rightarrow \frac{d y}{d x} = 1$

$∴ {(\frac{d y}{d x})}_{x = - 1} = 1$

Thus, if y = 2x + |x|, then ${(\frac{d y}{d x})}_{x = - 1} = 1$ and ${(\frac{d y}{d x})}_{x = 1} = 3$ .

If y = 2x + $|x|, then {(\frac{d y}{d x})}_{x = - 1} = 1 and {(\frac{d y}{d x})}_{x = 1} = 3 .$

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Question 3:

If f(x) = $|x^{2} - x|, then f' (2) =_________________.$

Answer:

$f (x) = |x^{2} - x| = \{\begin{cases} x^{2} - x, & x^{2} - x \geq 0 \\ - (x^{2} - x), & x^{2} - x < 0 \end{cases}$

$\Rightarrow f (x) = |x^{2} - x| = \{\begin{cases} x^{2} - x, & x (x - 1) \geq 0 \\ x - x^{2}, & x (x - 1) < 0 \end{cases}$

$\Rightarrow f (x) = |x^{2} - x| = \{\begin{cases} x^{2} - x, & - \infty < x \leq 0 or 1 \leq x < \infty \\ x - x^{2}, & 0 < x < 1 \end{cases}$

For x ≥ 1,

$f (x) = x^{2} - x$

$\Rightarrow f' (x) = 2 x - 1$

$∴ f' (2) = 2 \times 2 - 1 = 3$

If f(x) = $|x^{2} - x|, then f' (2) = 3 .$

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Question 4:

If y = sinx^o and $\frac{d y}{d x}$ = k cos x^o , then k = ________________.

Answer:

$y = \sin x °$

$\Rightarrow y = \sin (\frac{π}{180} x) (180 ° = π radians)$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} \sin (\frac{π}{180} x)$

$\Rightarrow \frac{d y}{d x} = \cos (\frac{π}{180} x) \times \frac{d}{d x} (\frac{π}{180} x)$

$\Rightarrow \frac{d y}{d x} = \cos x ° \times \frac{π}{180}$

$\Rightarrow \frac{d y}{d x} = \frac{π}{180} \cos x °$

Comparing with $\frac{d y}{d x} = k \cos x °$ , we get

$k = \frac{π}{180}$

If y = sinx^o and $\frac{d y}{d x}$ = k cosx^o , then k = $\frac{π}{180}$ .

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Question 5:

If f(x) = e^xg(x), g(0) = 2, g'(0) = 1, then f'(0) = __________________.

Answer:

$f (x) = e^{x} g (x)$

Differentiating both sides with respect to x, we get

$\frac{d}{d x} f (x) = \frac{d}{d x} [e^{x} g (x)]$

$\Rightarrow f' (x) = e^{x} \times \frac{d}{d x} g (x) + g (x) \times \frac{d}{d x} e^{x}$

$\Rightarrow f' (x) = e^{x} \times g' (x) + g (x) \times e^{x}$

Putting x = 0, we get

$f' (0) = e^{0} \times g' (0) + g (0) \times e^{0}$

$\Rightarrow f' (0) = 1 \times 1 + 2 \times 1$ [g(0) = 2, g'(0) = 1 and e⁰ = 1]

$\Rightarrow f' (0) = 1 + 2 = 3$

Thus, the value of $f' (0)$ is 3.

If f(x) = e^xg(x), g(0) = 2, g'(0) = 1, then f'(0) = ____3____.

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Question 6:

If f(x) = 3 $|x + 2|$ , then f '(-3) = ___________________.

Answer:

We know

$|x + 2| = \{\begin{cases} x + 2, & x \geq - 2 \\ - (x + 2), & x < - 2 \end{cases}$

$∴ f (x) = 3 |x + 2| = \{\begin{cases} 3 (x + 2), & x \geq - 2 \\ - 3 (x + 2), & x < - 2 \end{cases}$

$\Rightarrow f' (x) = \{\begin{cases} 3, & x \geq - 2 \\ - 3, & x < - 2 \end{cases}$

For x < −2, $f' (x) = - 3$

$∴ f' (- 3) = - 3$

If f(x) = 3 $|x + 2|$ , then f '(−3) = ____−3____.

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Question 7:

If f(1) = 3, f'(2) = 1, then $\frac{d}{d x} \{In f (e^{x} + 2 x)\} =____________________________.$

Answer:

Disclaimer: The solution is provided for the following question.

If f(1) = 3, f'(1) = 1, then $\frac{d}{d x} \{In f (e^{x} + 2 x)\} =____________________________.$

Solution:

$\frac{d}{d x} \{\ln f (e^{x} + 2 x)\} = \frac{1}{f (e^{x} + 2 x)} \times \frac{d}{d x} f (e^{x} + 2 x) = \frac{1}{f (e^{x} + 2 x)} \times f' (e^{x} + 2 x) \frac{d}{d x} (e^{x} + 2 x) = \frac{1}{f (e^{x} + 2 x)} \times f' (e^{x} + 2 x) \times (e^{x} + 2)$

Putting x = 0, we get

${[\frac{d}{d x} \{\ln f (e^{x} + 2 x)\}]}_{x = 0}$

$= \frac{1}{f (e^{0} + 2 \times 0)} \times f' (e^{0} + 2 \times 0) \times (e^{0} + 2)$

$= \frac{1}{f (1)} \times f' (1) \times (3) (e^{0} = 1)$

$= \frac{1}{3} \times 1 \times 3 [f (1) = 3, f' (1) = 1]$

= 1

If f(1) = 3, f'(1) = 1, then $\frac{d}{d x} \{In f (e^{x} + 2 x)\} =$ $1$ .

Page No 10.121:

Question 8:

If f(x) = x $|x|$ , then f '(x) = _________________.

Answer:

We know

$|x| = \{\begin{cases} x, & x \geq 0 \\ - x, & x < 0 \end{cases}$

$∴ f (x) = x |x| = \{\begin{cases} x^{2}, & x \geq 0 \\ - x^{2}, & x < 0 \end{cases}$

$\Rightarrow f' (x) = \{\begin{cases} 2 x, & x \geq 0 \\ - 2 x, & x < 0 \end{cases}$

Thus, $f' (x) = 2 x$ when x ≥ 0 and $f' (x) = - 2 x$ when x < 0.

If f(x) = $x |x|$ , then f '(x) = 2x when x ≥ 0 and −2x when x < 0.

Page No 10.121:

Question 9:

If f(x) = $|x - 1| + |x - 3|$ , then f '(2) = ______________________.

Answer:

We have

$|x - 1| = \{\begin{cases} x - 1, & x \geq 1 \\ - (x - 1), & x < 1 \end{cases}$

$|x - 3| = \{\begin{cases} x - 3, & x \geq 3 \\ - (x - 3), & x < 3 \end{cases}$

$∴ f (x) = |x - 1| + |x - 3| = \{\begin{cases} - (x - 1) - (x - 3), & x < 1 \\ (x - 1) - (x - 3), & 1 \leq x < 3 \\ (x - 1) + (x - 3), & x \geq 3 \end{cases}$

$\Rightarrow f (x) = |x - 1| + |x - 3| = \{\begin{cases} - 2 x + 4, & x < 1 \\ 2, & 1 \leq x < 3 \\ 2 x - 4, & x \geq 3 \end{cases}$

For 1 ≤ x < 3, f(x) = 2

∴ $f' (x) = 0$ , for 1 ≤ x < 3

$\Rightarrow f' (2) = 0$

Thus, the value of f '(2) is 0.

If f(x) = $|x - 1| + |x - 3|$ , then f '(2) = ___0___.

Page No 10.121:

Question 10:

If f(x) = $|c os x - \sin x|, then f' (\frac{π}{3}) =___________________.$

Answer:

For $\frac{π}{4} < x \leq \frac{π}{2}$ ,

$\cos x < \sin x$

$\Rightarrow \cos x - \sin x < 0$

$∴ f (x) = |\cos x - \sin x| = - (\cos x - \sin x)$

$\Rightarrow f (x) = - \cos x + \sin x$

Differentiating both sides with respect to x, we get

$\frac{d}{d x} f (x) = \frac{d}{d x} (- \cos x) + \frac{d}{d x} (\sin x)$

$\Rightarrow f' (x) = - (- \sin x) + \cos x$

$\Rightarrow f' (x) = \sin x + \cos x$

$∴ f' (\frac{π}{3}) = \sin \frac{π}{3} + \cos \frac{π}{3} = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2}$ $(\frac{π}{4} < \frac{π}{3} \leq \frac{π}{2})$

If f(x) = $|\cos x - \sin x|, then f' (\frac{π}{3}) = \frac{\sqrt{3} + 1}{2} .$

Page No 10.121:

Question 11:

If f(x) = $|\cos x|, then f' (\frac{π}{4}) =______________________.$

Answer:

For $0 \leq x < \frac{π}{2}$ ,

cosx > 0

$∴ f (x) = |\cos x| = \cos x$

$\Rightarrow f' (x) = - \sin x$

$∴ f' (\frac{π}{4}) = - \sin \frac{π}{4} = - \frac{1}{\sqrt{2}}$

If f(x) = $|\cos x|, then f' (\frac{π}{4}) = - \frac{1}{\sqrt{2}} .$

Page No 10.121:

Question 12:

The derivative of x² with respect to x³ is __________________.

Answer:

Let u(x) = x² and v(x) =x³.

$u (x) = x^{2}$

$\Rightarrow \frac{d u}{d x} = 2 x$

$v (x) = x^{3}$

$\Rightarrow \frac{d v}{d x} = 3 x^{2}$

$∴ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

$\Rightarrow \frac{d u}{d v} = \frac{2 x}{3 x^{2}} = \frac{2}{3 x}$

Thus, the derivative of x² with respect to x³ is $\frac{2}{3 x}$ .

The derivative of x² with respect to x³ is $\frac{2}{3 x}$ .

Page No 10.121:

Question 13:

For the curve $\sqrt{x} + \sqrt{y} = 1, \frac{d y}{d x} at (\frac{1}{4}, \frac{1}{4}) is______________________.$

Answer:

$\sqrt{x} + \sqrt{y} = 1$ (Given)

Differentiating both sides with respect to x, we get

$\frac{d}{d x} \sqrt{x} + \frac{d}{d x} \sqrt{y} = \frac{d}{d x} (1)$

$\Rightarrow \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = 0$

$\Rightarrow \frac{1}{2 \sqrt{y}} \frac{d y}{d x} = - \frac{1}{2 \sqrt{x}}$

$\Rightarrow \frac{d y}{d x} = - \frac{\sqrt{y}}{\sqrt{x}} = - \sqrt{\frac{y}{x}}$

$∴ {(\frac{d y}{d x})}_{(\frac{π}{4}, \frac{π}{4})} = - \sqrt{\frac{\frac{1}{4}}{\frac{1}{4}}}$

$\Rightarrow {(\frac{d y}{d x})}_{(\frac{π}{4}, \frac{π}{4})} = - 1$

Thus, the value of $\frac{d y}{d x}$ at $(\frac{π}{4}, \frac{π}{4})$ is −1.

For the curve $\sqrt{x} + \sqrt{y} = 1, \frac{d y}{d x} at (\frac{1}{4}, \frac{1}{4}) is - 1 .$

Page No 10.121:

Question 14:

If f(x) = $|\sin x|, then f' (- \frac{π}{4}) =____________________.$

Answer:

For $- \frac{π}{2} < x < 0$ ,

$\sin x < 0$

$∴ f (x) = |\sin x| = - \sin x$

$\Rightarrow f' (x) = - \cos x$

$∴ f' (- \frac{π}{4}) = - \cos (- \frac{π}{4}) = - \cos \frac{π}{4} = - \frac{1}{\sqrt{2}} [\cos (- θ) = \cos θ]$

If f(x) = $|\sin x|, then f' (- \frac{π}{4}) = - \frac{1}{\sqrt{2}} .$

Page No 10.121:

Question 15:

If f(x) = $|\sin x - \cos x|, then f' (\frac{π}{6}) =________________________.$

Answer:

For $0 < x < \frac{π}{4}$ ,

$\sin x < \cos x$

$\Rightarrow \sin x - \cos x < 0$

$∴ f (x) = |\sin x - \cos x| = - (\sin x - \cos x)$

$\Rightarrow f (x) = - \sin x + \cos x$

Differentiating both sides with respect to x, we get

$f' (x) = - \cos x - \sin x$

$∴ f' (\frac{π}{6}) = - \cos \frac{π}{6} - \sin \frac{π}{6} = - \frac{\sqrt{3}}{2} - \frac{1}{2} = - \frac{1}{2} (\sqrt{3} + 1)$

If f(x) = $|\sin x - \cos x|, then f' (\frac{π}{6}) = - \frac{1}{2} (\sqrt{3} + 1) .$

Page No 10.121:

Question 16:

If y = tan x^o, then ${(\frac{d y}{d x})}_{x = 45^{o}}$ = ________________________.

Answer:

$y = \tan x °$

$\Rightarrow y = \tan (\frac{π}{180} x) (180 ° = π radians)$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} \tan (\frac{π}{180} x)$

$\Rightarrow \frac{d y}{d x} = \sec^{2} (\frac{π}{180} x) \times \frac{d}{d x} (\frac{π}{180} x)$

$\Rightarrow \frac{d y}{d x} = \frac{π}{180} \sec^{2} (\frac{π}{180} x)$

$\Rightarrow \frac{d y}{d x} = \frac{π}{180} \sec^{2} x °$

$∴ {(\frac{d y}{d x})}_{x = 45 °} = \frac{π}{180} \sec^{2} 45 °$

$\Rightarrow {(\frac{d y}{d x})}_{x = 45 °} = \frac{π}{180} \times {(\sqrt{2})}^{2} = \frac{π}{90}$

If y = tanxº, then ${(\frac{d y}{d x})}_{x = 45^{o}}$ = $\frac{π}{90}$ .

Page No 10.121:

Question 17:

If y = sin^-1(e^x) + cos^-1(e^x), then $\frac{d y}{d x}$ = ____________________.

Answer:

$y = \sin^{- 1} (e^{x}) + \cos^{- 1} (e^{x})$

$\Rightarrow y = \frac{π}{2}$ $(\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2})$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} (\frac{π}{2})$

$\Rightarrow \frac{d y}{d x} = 0$

If y = sin⁻¹(e^x) + cos⁻¹(e^x), then $\frac{d y}{d x}$ = ____0____.

Page No 10.121:

Question 18:

If y = sin^-1(3x-4x³), $\frac{1}{2}$ < x < 1, then $\frac{d y}{d x}$ = ______________________.

Answer:

y = sin⁻¹(3x − 4x³)

Let x = sinθ.

$∴ y = \sin^{- 1} (3 \sin θ - 4 \sin^{3} θ)$

$\Rightarrow y = \sin^{- 1} (\sin 3 θ)$

Now,

$\frac{1}{2} < x < 1$

$\Rightarrow \frac{1}{2} < \sin θ < 1$

$\Rightarrow \frac{π}{6} < θ < \frac{π}{2}$

$\Rightarrow \frac{π}{2} < 3 θ < \frac{3 π}{2}$

$∴ y = \sin^{- 1} (\sin 3 θ)$

$\Rightarrow y = \sin^{- 1} [\sin (π - 3 θ)]$

$\Rightarrow y = π - 3 θ [\frac{π}{2} < 3 θ < \frac{3 π}{2} \Rightarrow - \frac{π}{2} < π - 3 θ < \frac{π}{2}]$

$\Rightarrow y = π - 3 \sin^{- 1} x$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 0 - 3 \times \frac{1}{\sqrt{1 - x^{2}}} \Rightarrow \frac{d y}{d x} = - \frac{3}{\sqrt{1 - x^{2}}}$

If y = sin⁻¹(3x − 4x³), $\frac{1}{2}$ < x < 1, then $\frac{d y}{d x}$ = $- \frac{3}{\sqrt{1 - x^{2}}}$ .

Page No 10.121:

Question 19:

If y = sec^-1 $(\frac{\sqrt{x} + 1}{\sqrt{x} - 1}) + \sin^{- 1} (\frac{\sqrt{x} - 1}{\sqrt{x} + 1}), then \frac{d y}{d x}$ is equal to ___________________.

Answer:

We know

$\sec^{- 1} a = \cos^{- 1} \frac{1}{a}$

$∴ \sec^{- 1} (\frac{\sqrt{x} + 1}{\sqrt{x} - 1}) = \cos^{- 1} (\frac{\sqrt{x} - 1}{\sqrt{x} + 1})$ .....(1)

Now,

$y = \sec^{- 1} (\frac{\sqrt{x} + 1}{\sqrt{x} - 1}) + \sin^{- 1} (\frac{\sqrt{x} - 1}{\sqrt{x} + 1})$

$\Rightarrow y = \cos^{- 1} (\frac{\sqrt{x} - 1}{\sqrt{x} + 1}) + \sin^{- 1} (\frac{\sqrt{x} - 1}{\sqrt{x} + 1})$ [Using (1)]

$\Rightarrow y = \frac{π}{2} (\sin^{- 1} a + \cos^{- 1} a = \frac{π}{2})$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = 0$

Thus, if $y = \sec^{- 1} (\frac{\sqrt{x} + 1}{\sqrt{x - 1}}) + \sin^{- 1} (\frac{\sqrt{x} - 1}{\sqrt{x} + 1})$ , then $\frac{d y}{d x} = 0$ .

If $y = \sec^{- 1} (\frac{\sqrt{x} + 1}{\sqrt{x - 1}}) + \sin^{- 1} (\frac{\sqrt{x} - 1}{\sqrt{x} + 1})$ , then $\frac{d y}{d x}$ is equal to ___0___.

Page No 10.121:

Question 20:

The derivative of cos x with respect to sin x is __________________.

Answer:

Let u(x) = cosx and v(x) = sinx.

$u (x) = \cos x$

$\Rightarrow \frac{d u}{d x} = \frac{d}{d x} \cos x$

$\Rightarrow \frac{d u}{d x} = - \sin x$      .....(1)

$v (x) = \sin x$

$\Rightarrow \frac{d v}{d x} = \frac{d}{d x} \sin x$

$\Rightarrow \frac{d v}{d x} = \cos x$    .....(2)

$∴ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

$\Rightarrow \frac{d u}{d v} = \frac{- \sin x}{\cos x}$        [From (1) and (2)]

$\Rightarrow \frac{d u}{d v} = - \tan x$

Thus, the derivative of cosx with respect to sinx is −tanx.

The derivative of cos x with respect to sin x is ___−tanx___.

Page No 10.121:

Question 21:

The derivative of log₁₀x with respect to x is ___________________.

Answer:

Let $y = \log_{10} x$ .

$y = \log_{10} x$

$\Rightarrow y = \frac{\log_{e} x}{\log_{e} 10} (\log_{b} a = \frac{\log_{c} a}{\log_{c} b})$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} (\frac{\log_{e} x}{\log_{e} 10})$

$\Rightarrow \frac{d y}{d x} = \frac{1}{\log_{e} 10} \times \frac{d}{d x} (\log_{e} x)$

$\Rightarrow \frac{d y}{d x} = \frac{1}{\log_{e} 10} \times \frac{1}{x}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{x \log_{e} 10}$

Thus, the derivative of $\log_{10} x$ with respect to x is $\frac{1}{x \log_{e} 10}$ .

The derivative of log₁₀x with respect to x is $\frac{1}{x \log_{e} 10}$ .

Page No 10.121:

Question 22:

$If \frac{d}{d x} (f (x)) = \frac{1}{1 + x^{2}}, then \frac{d}{d x} \{f (x^{3})\} =_________________________.$

Answer:

$\frac{d}{d x} f (x^{3}) = f' (x^{3}) \times \frac{d}{d x} x^{3}$

$\Rightarrow \frac{d}{d x} f (x^{3}) = f' (x^{3}) \times 3 x^{2}$      .....(1)

Now,

$\frac{d}{d x} f (x) = f' (x) = \frac{1}{1 + x^{2}}$           (Given)

Replacing x by x³, we get

$f' (x^{3}) = \frac{1}{1 + {(x^{3})}^{2}}$

$\Rightarrow f' (x^{3}) = \frac{1}{1 + x^{6}}$          .....(2)

From (1) and (2), we get

$\frac{d}{d x} f (x^{3}) = \frac{1}{1 + x^{6}} \times 3 x^{2}$

$\Rightarrow \frac{d}{d x} f (x^{3}) = \frac{3 x^{2}}{1 + x^{6}}$

$If \frac{d}{d x} (f (x)) = \frac{1}{1 + x^{2}}, then \frac{d}{d x} \{f (x^{3})\} = \frac{3 x^{2}}{1 + x^{6}} .$

Page No 10.121:

Question 23:

If y = cos (sin x²), then $\frac{d y}{d x} at x = \sqrt{\frac{π}{2}}$ is equal to ______________________.

Answer:

$y = \cos (\sin x^{2})$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} \cos (\sin x^{2})$

$\Rightarrow \frac{d y}{d x} = - \sin (\sin x^{2}) \times \frac{d}{d x} \sin x^{2}$

$\Rightarrow \frac{d y}{d x} = - \sin (\sin x^{2}) \times \cos x^{2} \times \frac{d}{d x} x^{2}$

$\Rightarrow \frac{d y}{d x} = - \sin (\sin x^{2}) \times \cos x^{2} \times 2 x$

$\Rightarrow \frac{d y}{d x} = - 2 x \cos x^{2} \sin (\sin x^{2})$

Putting $x = \sqrt{\frac{π}{2}}$ , we get

${(\frac{d y}{d x})}_{x = \sqrt{\frac{π}{2}}} = - 2 \times \sqrt{\frac{π}{2}} \times \cos {(\sqrt{\frac{π}{2}})}^{2} \sin [\sin {(\sqrt{\frac{π}{2}})}^{2}]$

$\Rightarrow {(\frac{d y}{d x})}_{x = \sqrt{\frac{π}{2}}} = - 2 \times \sqrt{\frac{π}{2}} \times \cos \frac{π}{2} \times \sin (\sin \frac{π}{2})$

$\Rightarrow {(\frac{d y}{d x})}_{x = \sqrt{\frac{π}{2}}} = - 2 \times \sqrt{\frac{π}{2}} \times 0 \times \sin 1$

$\Rightarrow {(\frac{d y}{d x})}_{x = \sqrt{\frac{π}{2}}} = 0$

Thus, $\frac{d y}{d x}$ at $x = \sqrt{\frac{π}{2}}$ is 0.

If y = cos (sin x²), then $\frac{d y}{d x} at x = \sqrt{\frac{π}{2}}$ is equal to ___0___.

Page No 10.121:

Question 24:

If y = log $|x|, x \neq 0, then \frac{d y}{d x}$ = ___________________.

Answer:

For $y = \log |x|$ to be defined,

$|x| \neq 0$

$\Rightarrow x \neq 0$

Now,

$y = \log |x| = \{\begin{cases} \log x, & x > 0 \\ \log (- x), & x < 0 \end{cases}$

$∴ \frac{d y}{d x} = \{\begin{cases} \frac{1}{x}, & x > 0 \\ \frac{1}{(- x)} \times (- 1), & x < 0 \end{cases}$

$\Rightarrow \frac{d y}{d x} = \{\begin{cases} \frac{1}{x}, & x > 0 \\ \frac{1}{x}, & x < 0 \end{cases}$

$∴ \frac{d y}{d x} = \frac{1}{x}, x \neq 0$

If y = log $|x|, x \neq 0, then \frac{d y}{d x}$ = $\frac{1}{x}$ .

Page No 10.121:

Question 25:

If f(x) = ax² + bx + c, then f '(1) + f '(4) - f '(5) is equal to _____________________.

Answer:

$f (x) = a x^{2} + b x + c$

$∴ f' (x) = \frac{d}{d x} (a x^{2} + b x + c)$

$\Rightarrow f' (x) = 2 a x + b$

$∴ f' (1) + f' (4) - f' (5)$

$= (2 a \times 1 + b) + (2 a \times 4 + b) - (2 a \times 5 + b)$

$= 2 a + b + 8 a + b - 10 a - b$

$= b$

Thus, the value of $f' (1) + f' (4) - f' (5)$ is b.

If f(x) = ax² + bx + c, then f '(1) + f '(4) − f '(5) is equal to ___b___.

Page No 10.121:

Question 26:

If f '(1) = 2 and g' $(\sqrt{2})$ = 4, then the derivative of f(tan x) with respect of g(secx) at x = $\frac{π}{4}$ is equal to ______________.

Answer:

Let u(x) = f(tanx) and v(x) = g(secx).

$u (x) = f (\tan x)$

$\Rightarrow \frac{d u}{d x} = \frac{d}{d x} f (\tan x)$

$\Rightarrow \frac{d u}{d x} = f' (\tan x) \frac{d}{d x} \tan x$

$\Rightarrow \frac{d u}{d x} = f' (\tan x) \times \sec^{2} x$ .....(1)

$v (x) = g (\sec x)$

$\Rightarrow \frac{d v}{d x} = \frac{d}{d x} g (\sec x)$

$\Rightarrow \frac{d v}{d x} = g' (\sec x) \frac{d}{d x} \sec x$

$\Rightarrow \frac{d v}{d x} = g' (\sec x) \times \sec x \tan x$ .....(2)

$∴ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

$\Rightarrow \frac{d u}{d v} = \frac{\sec^{2} x \times f' (\tan x)}{\sec x \tan x \times g' (\sec x)}$

$\Rightarrow \frac{d u}{d v} = \frac{\sec x \times f' (\tan x)}{\tan x \times g' (\sec x)}$

Putting $x = \frac{π}{4}$ , we get

${(\frac{d u}{d v})}_{x = \frac{π}{4}} = \frac{\sec \frac{π}{4} \times f' (\tan \frac{π}{4})}{\tan \frac{π}{4} \times g' (\sec \frac{π}{4})}$

$\Rightarrow {(\frac{d u}{d v})}_{x = \frac{π}{4}} = \frac{\sqrt{2} \times f' (1)}{1 \times g' (\sqrt{2})}$

$\Rightarrow {(\frac{d u}{d v})}_{x = \frac{π}{4}} = \frac{\sqrt{2} \times 2}{1 \times 4}$

$\Rightarrow {(\frac{d u}{d v})}_{x = \frac{π}{4}} = \frac{1}{\sqrt{2}}$

Thus, the derivative of f(tan x) with respect of g(secx) at x = $\frac{π}{4}$ is $\frac{1}{\sqrt{2}}$ .

If f '(1) = 2 and g' $(\sqrt{2})$ = 4, then the derivative of f(tan x) with respect of g(secx) at x = $\frac{π}{4}$ is equal to $\frac{1}{\sqrt{2}}$ .

Page No 10.122:

Question 1:

If f (x) = log_e (log_e x), then write the value of f' (e).

Answer:

$We have, f (x) = \log_{e} (\log_{e} x)$
Differentiating with respect to x,
$f' (x) = \frac{1}{\log_{e} x} \frac{d}{d x} (\log_{e} x) \Rightarrow f' (x) = \frac{1}{\log_{e} x} (\frac{1}{x}) \Rightarrow f' (e) = \frac{1}{\log_{e} e} (\frac{1}{e}) [∵ x = e] \Rightarrow f' (e) = \frac{1}{e} [∵ \log_{e} e = 1]$

Page No 10.122:

Question 2:

If $f (x) = x + 1$ , then write the value of $\frac{d}{d x} (f o f) (x)$ .

Answer:

$We have, f (x) = x + 1 Now, (f o f) (x) = f (f (x)) \Rightarrow (f o f) (x) = f (x + 1) \Rightarrow (f o f) (x) = (x + 1) + 1 \Rightarrow (f o f) = x + 2$

$\Rightarrow \frac{d}{d x} \{(f o f) (x)\} = \frac{d}{d x} (x) + \frac{d}{d x} (2) \Rightarrow \frac{d}{d x} \{(f o f) (x)\} = 1 + 0 \Rightarrow \frac{d}{d x} \{(f o f) (x)\} = 1$

Page No 10.122:

Question 3:

If $f' (1) = 2 and y = f (\log_{e} x), find \frac{d y}{d x} a t x = e$ .

Answer:

$We have, f' (1) = 2 and y = f (\log_{e} x)$
Differentiate it with respect to x,
$\frac{d y}{d x} = f' (\log_{e} x) \times \frac{d}{d x} (\log_{e} x) \Rightarrow \frac{d y}{d x} = f' (\log_{e} x) (\frac{1}{x}) \Rightarrow \frac{d y}{d x} = f' (\log_{e} e) (\frac{1}{e}) [∵ x = e] \Rightarrow \frac{d y}{d x} = f' (1) (\frac{1}{e}) [∵ \log_{e} e = 1] \Rightarrow \frac{d y}{d x} = \frac{2}{e} [∵ f' (1) = 2]$

Page No 10.122:

Question 4:

If $f (1) = 4, f' (1) = 2$ , find the value of the derivative of $\log (f (e^{x}))$ w.r. to x at the point x = 0.

Answer:

$We have, f (1) = 4 and f' (1) = 2 Let y = \log \{f (e^{x})\}$

$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} [\log \{f (e^{x})\}] \Rightarrow \frac{d y}{d x} = \frac{1}{f (e^{x})} \times \frac{d}{d x} \{f (e^{x})\} \Rightarrow \frac{d y}{d x} = \frac{1}{f (e^{x})} \times f' (e^{x}) \times \frac{d}{d x} (e^{x}) \Rightarrow \frac{d y}{d x} = \frac{e^{x} f' (e^{x})}{f (e^{x})} Putting x = 0, we get, \frac{d y}{d x} = \frac{e^{0} f' (e^{0})}{f (e^{0})} \Rightarrow \frac{d y}{d x} = \frac{1 f' (1)}{f (1)} \Rightarrow \frac{d y}{d x} = \frac{2}{4} [∵ f' (1) = 2 and f (1) = 4] \Rightarrow \frac{d y}{d x} = \frac{1}{2}$

Page No 10.122:

Question 5:

If $f' (x) = \sqrt{2 x^{2} - 1} and y = f (x^{2})$ , then find $\frac{d y}{d x} a t x = 1$ .

Answer:

$We have, f' (x) = \sqrt{2 x^{2} - 1} and y = f (x^{2})$

$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} f (x^{2}) \Rightarrow \frac{d y}{d x} = f' (x^{2}) \frac{d}{d x} (x^{2}) \Rightarrow \frac{d y}{d x} = f' (x^{2}) \times 2 x \Rightarrow \frac{d y}{d x} = 2 x f' (x^{2}) Putting x = 1, we get, \frac{d y}{d x} = 2 (1) f' (1^{2}) \Rightarrow \frac{d y}{d x} = 2 \times f' (1) \Rightarrow \frac{d y}{d x} = 2 \times 1 [∵ f' (1) = \sqrt{2 {(1)}^{2} - 1} = \sqrt{2 - 1} = 1] \Rightarrow \frac{d y}{d x} = 2$

Page No 10.122:

Question 6:

Let g (x) be the inverse of an invertible function f (x) which is derivable at x = 3. If f (3) = 9 and f' (3) = 9, write the value of g' (9).

Answer:

$We have, f (3) = 9, f' (3) = 9 and g (x) = f^{- 1} (x) \Rightarrow (g o f) (x) = x \Rightarrow g \{f (x)\} = x$

$\Rightarrow \frac{d}{d x} [g \{f (x)\}] = 1 \Rightarrow g' \{f (x)\} \frac{d}{d x} \{f (x)\} = 1 \Rightarrow g' \{f (x)\} \times f' (x) = 1 Puting x = 3, we get, g' \{f (3)\} \times f' (3) = 1 \Rightarrow g' (9) \times 9 = 1 [∵ f (3) = 9, f' (3) = 9] \Rightarrow g' (9) = \frac{1}{9}$

Page No 10.122:

Question 7:

If $y = \sin^{- 1} (\sin x), - \frac{π}{2} \leq x \leq \frac{π}{2}$ . Then, write the value of $\frac{d y}{d x} for x \in (- \frac{π}{2}, \frac{π}{2}) .$

Answer:

$We have, y = \sin^{- 1} (\sin x) \Rightarrow y = x [∵ \sin^{- 1} (\sin x) = x, if x \in [- \frac{π}{2}, \frac{π}{2}]]$

$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} (x) \Rightarrow \frac{d y}{d x} = 1$

Page No 10.122:

Question 8:

If $\frac{π}{2} \leq x \leq \frac{3 π}{2} and y = \sin^{- 1} (\sin x), find \frac{d y}{d x} .$

Answer:

$We have, y = \sin^{- 1} (\sin x) \Rightarrow y = π - x [∵ \sin^{- 1} (\sin x) = π - x, if x \in [\frac{π}{2}, \frac{3 π}{2}]]$

$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} (π - x) \Rightarrow \frac{d y}{d x} = 0 - 1 \Rightarrow \frac{d y}{d x} = - 1$

Page No 10.122:

Question 9:

If $π \leq x \leq 2 π and y = \cos^{- 1} (\cos x), find \frac{d y}{d x}$ .

Answer:

$We have, y = \cos^{- 1} (\cos x) \Rightarrow y = 2 π - x [∵ \cos^{- 1} (\cos x) = 2 π - x, if x \in [π, 2 π]]$

$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} (2 π - x) \Rightarrow \frac{d y}{d x} = 0 - 1 \Rightarrow \frac{d y}{d x} = - 1$

Page No 10.122:

Question 10:

If $y = \sin^{- 1} (\frac{2 x}{1 + x^{2}})$ , write the value of $\frac{d y}{d x} for x > 1$ .

Answer:

$We have, y = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) Putting x = \tan θ \Rightarrow 1 < \tan θ < \infty \Rightarrow \frac{π}{4} < θ < \frac{π}{2} \Rightarrow \frac{π}{2} < 2 θ < π ∴ y = \sin^{- 1} (\sin 2 θ) \Rightarrow y = \sin^{- 1} \{\sin (π - 2 θ)\} \Rightarrow y = π - 2 θ \Rightarrow y = π - 2 \tan^{- 1} x \Rightarrow \frac{d y}{d x} = 0 - \frac{2}{1 + x^{2}} \Rightarrow \frac{d y}{d x} = - \frac{2}{1 + x^{2}}$

Page No 10.122:

Question 11:

If $f (0) = f (1) = 0, f' (1) = 2 and y = f (e^{x}) e^{f (x)}$ , write the value of $\frac{d y}{d x} at x = 0$ .

Answer:

$We have, f (0) = f (1) = 0, f' (1) = 2 and, y = f (e^{x}) e^{f (x)}$

$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} [f (e^{x}) \times e^{f (x)}] \Rightarrow \frac{d y}{d x} = f (e^{x}) \frac{d}{d x} e^{f (x)} + e^{f (x)} \frac{d}{d x} f (e^{x}) [Using product rule] \Rightarrow \frac{d y}{d x} = f (e^{x}) \times e^{f (x)} \frac{d}{d x} f (x) + e^{f (x)} \times f' (e^{x}) \frac{d}{d x} (e^{x}) \Rightarrow \frac{d y}{d x} = f (e^{x}) \times e^{f (x)} \times f' (x) + e^{f (x)} \times f' (e^{x}) \times e^{x} Putting x = 0, we get, \frac{d y}{d x} = f (e^{0}) \times e^{f (0)} \times f' (0) + e^{f (0)} \times f' (e^{0}) \times e^{0} \Rightarrow \frac{d y}{d x} = f (1) e^{f (0)} \times f' (0) + e^{f (0)} \times f' (1) \times 1 \Rightarrow \frac{d y}{d x} = 0 \times e^{0} \times f' (0) + e^{0} \times 2 \times 1 [∵ f (x) = f (1) = 0 and f' (1) = 2] \Rightarrow \frac{d y}{d x} = 0 + 1 \times 2 \times 1 \Rightarrow \frac{d y}{d x} = 2$

Page No 10.122:

Question 12:

If $y = x |x|$ , find $\frac{d y}{d x} for x < 0$ .

Answer:

$We have, y = x |x| \Rightarrow y = x (- x) (∵ x < 0) \Rightarrow y = - x^{2}$

$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} (- x^{2}) \Rightarrow \frac{d y}{d x} = - 2 x$

Page No 10.122:

Question 13:

If $y = \sin^{- 1} x + \cos^{- 1} x$ , find $\frac{d y}{d x}$ .

Answer:

$We have, y = \sin^{- 1} x + \cos^{- 1} x \Rightarrow y = \frac{π}{2} [∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}]$

$\Rightarrow \frac{d y}{d x} = 0$

Page No 10.122:

Question 14:

If $x = a (θ + \sin θ), y = a (1 + \cos θ), find \frac{d y}{d x}$ .

Answer:

$We have, x = a (θ + \sin θ) and y = a (1 + \cos θ) \Rightarrow \frac{d x}{d θ} = a \{\frac{d}{d θ} (θ) + \frac{d}{d θ} (\sin θ)\} and \frac{d y}{d θ} = a (0 - \sin θ) \Rightarrow \frac{d x}{d θ} = a (1 + \cos θ) and \frac{d y}{d θ} = - a \sin θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{- a \sin θ}{a (1 + \cos θ)} = \frac{- 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}} = - \tan \frac{θ}{2}$

Page No 10.122:

Question 15:

If $- \frac{π}{2} < x < 0 and y = \tan^{- 1} \sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}}, find \frac{d y}{d x}$ .

Answer:

$We have, y = \tan^{- 1} \sqrt{\frac{1 - \cos 2 x}{1 + \cos 2 x}} \Rightarrow y = \tan^{- 1} \sqrt{\frac{2 \sin^{2} x}{2 \cos^{2} x}} \Rightarrow y = \tan^{- 1} \sqrt{\tan^{2} x} \Rightarrow y = \tan^{- 1} (\tan x) [∵ \tan^{- 1} (\tan x) = - x, if x \in (- \frac{π}{2}, 0)] \Rightarrow y = - x$

$\Rightarrow \frac{d y}{d x} = - 1$

Page No 10.122:

Question 16:

If $y = x^{x}, find \frac{d y}{d x} a t x = e$ .

Answer:

$We have, y = x^{x} . . . (i)$
Taking log on both sides,
$\log y = \log x^{x} \Rightarrow \log y = x \log x$

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = x \frac{d}{d x} (\log x) + \log x \frac{d}{d x} (x) \Rightarrow \frac{1}{y} \frac{d y}{d x} = x (\frac{1}{x}) + \log x (1) \Rightarrow \frac{1}{y} \frac{d y}{d x} = 1 + \log x \Rightarrow \frac{d y}{d x} = y (1 + \log x) \Rightarrow \frac{d y}{d x} = x^{x} (1 + \log x) [using equation (i)] Puting x = e, we get, \frac{d y}{d x} = e^{e} (1 + \log_{e} e) \Rightarrow \frac{d y}{d x} = e^{e} (1 + 1) [∵ \log_{e} e = 1] \Rightarrow \frac{d y}{d x} = 2 e^{e}$

Page No 10.123:

Question 17:

If $y = \tan^{- 1} (\frac{1 - x}{1 + x}), find \frac{d y}{d x}$ .

Answer:

$We have, y = \tan^{- 1} (\frac{1 - x}{1 + x})$

$\Rightarrow \frac{d y}{d x} = \frac{1}{1 + {(\frac{1 - x}{1 + x})}^{2}} \frac{d}{d x} (\frac{1 - x}{1 + x}) \Rightarrow \frac{d y}{d x} = \frac{{(1 + x)}^{2}}{1 + x^{2} + 2 x + 1 + x^{2} - 2 x} [\frac{(1 + x) \frac{d}{d x} (1 - x) - (1 - x) \frac{d}{d x} (1 + x)}{{(1 + x)}^{2}}] [using quotient rule] \Rightarrow \frac{d y}{d x} = \frac{{(1 + x)}^{2}}{2 x^{2} + 2} [\frac{(1 + x) (- 1) - (1 - x) (1)}{(1 + x)}] \Rightarrow \frac{d y}{d x} = \frac{{(1 + x)}^{2}}{2 (x^{2} + 1)} (\frac{- x - 1 - 1 + x}{{(1 + x)}^{2}}) \Rightarrow \frac{d y}{d x} = \frac{{(1 + x)}^{2}}{2 (x^{2} + 1)} \times \frac{- 2}{{(1 + x)}^{2}} \Rightarrow \frac{d y}{d x} = - \frac{1}{x^{2} + 1}$

Page No 10.123:

Question 18:

If $y = \log_{a} x, find \frac{d y}{d x} .$

Answer:

$We have, y = \log_{a} x \Rightarrow y = \frac{\log x}{\log a} [∵ \log_{a} b = \frac{\log b}{\log a}]$

$\Rightarrow \frac{d y}{d x} = \frac{1}{\log a} \frac{d}{d x} (\log x) \Rightarrow \frac{d y}{d x} = \frac{1}{\log a} (\frac{1}{x}) \Rightarrow \frac{d y}{d x} = \frac{1}{x \log a}$

Page No 10.123:

Question 19:

If $y = \log \sqrt{\tan x}, write \frac{d y}{d x} .$

Answer:

$We have, y = \log \sqrt{\tan x} \Rightarrow y = \log {(\tan x)}^{\frac{1}{2}} \Rightarrow y = \frac{1}{2} \log \tan x [∵ \log a^{b} = b \log a]$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2} \times \frac{1}{\tan x} \frac{d}{d x} (\tan x) \Rightarrow \frac{d y}{d x} = \frac{1}{2} \times \frac{1}{\tan x} (\sec^{2} x) \Rightarrow \frac{d y}{d x} = \frac{1}{2 \frac{\sin x}{\cos x} \times \cos^{2} x} \Rightarrow \frac{d y}{d x} = \frac{1}{2 \sin x \cos x} \Rightarrow \frac{d y}{d x} = \frac{1}{\sin 2 x} \Rightarrow \frac{d y}{d x} = cosec 2 x$

Page No 10.123:

Question 20:

If $y = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), find \frac{d y}{d x}$ .

Answer:

$We have, y = \sin^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) \Rightarrow y = \frac{π}{2} [∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}]$

$\Rightarrow \frac{d y}{d x} = 0$

Page No 10.123:

Question 21:

If $y = \sec^{- 1} (\frac{x + 1}{x - 1}) + \sin^{- 1} (\frac{x - 1}{x + 1})$ , then write the value of $\frac{d y}{d x} .$

Answer:

$We have, y = \sec^{- 1} (\frac{x + 1}{x - 1}) + \sin^{- 1} (\frac{x - 1}{x + 1}) \Rightarrow y = \cos^{- 1} (\frac{x - 1}{x + 1}) + \sin^{- 1} (\frac{x - 1}{x + 1}) [∵ \sec^{- 1} x = \cos^{- 1} (\frac{1}{x})] \Rightarrow y = \frac{π}{2} [∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}]$

$\Rightarrow \frac{d y}{d x} = 0$

Page No 10.123:

Question 22:

If $|x| < 1 and y = 1 + x + x^{2} + . . .$ to ∞, then find the value of $\frac{d y}{d x}$ .

Answer:

$We have, y = 1 + x + x^{2} + . . . . to \infty \Rightarrow y = \frac{1}{1 - x} [\begin{matrix} ∵ It is a G . P with first term 1 \\ and common ratio x \end{matrix}]$

$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} (\frac{1}{1 - x}) \Rightarrow \frac{d y}{d x} = - \frac{1}{{(1 - x)}^{2}} \frac{d}{d x} (1 - x) \Rightarrow \frac{d y}{d x} = - \frac{1}{{(1 - x)}^{2}} (- 1) \Rightarrow \frac{d y}{d x} = \frac{1}{{(1 - x)}^{2}}$

Page No 10.123:

Question 23:

If $u = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) and v = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ , where $- 1 < x < 1$ , then write the value of $\frac{d u}{d v}$ .

Answer:

$We have, u = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) and v = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) \Rightarrow \frac{d u}{d x} = \frac{2}{1 + x^{2}} and \frac{d v}{d x} = \frac{2}{1 + x^{2}} ∴ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{2}{1 + x^{2}} \times \frac{1 + x^{2}}{2} = 1$

Page No 10.123:

Question 24:

If $f (x) = \log \{\frac{u (x)}{v (x)}\}, u (1) = v (1) and u' (1) = v' (1) = 2$ , then find the value of f' (1).

Answer:

$We have, f (x) = \log \{\frac{u (x)}{v (x)}\} and, u (1) = v (1), u' (1) = v' (1) = 2 . . . (i)$

$\Rightarrow f' (x) = \frac{d}{d x} [\log \{\frac{u (x)}{v (x)}\}] \Rightarrow f' (x) = \frac{1}{[\frac{u (x)}{v (x)}]} \times \frac{d}{d x} [\frac{u (x)}{v (x)}] \Rightarrow f' (x) = \frac{v (x)}{u (x)} \times [\frac{v (x) \frac{d}{d x} \{u (x)\} - u (x) \frac{d}{d x} \{v (x)\}}{{\{v (x)\}}^{2}}] \Rightarrow f' (x) = \frac{v (x)}{u (x)} \times [\frac{v (x) \times u' (x) - u (x) \times v' (x)}{{\{v (x)\}}^{2}}] Putting x = 1, we get, f' (1) = \frac{v (1)}{u (1)} \times [\frac{v (1) \times u' (1) - u (1) \times v' (1)}{{\{v (1)\}}^{2}}] \Rightarrow f' (1) = 1 \times [\frac{u (1) \times 2 - u (1) \times 2}{{\{u (1)\}}^{2}}] [Using eqn (1)] \Rightarrow f' (1) = [\frac{0}{{\{u (1)\}}^{2}}] \Rightarrow f' (1) = 0$

Page No 10.123:

Question 25:

If $y = \log |3 x|, x \neq 0, find \frac{d y}{d x} .$

Answer:

$We have, y = \log |3 x|$
$\Rightarrow \frac{d y}{d x} = \frac{d}{d x} (\log |3 x|) \Rightarrow \frac{d y}{d x} = \frac{1}{3 x} \frac{d}{d x} (3 x) \Rightarrow \frac{d y}{d x} = \frac{1}{3 x} (3) \Rightarrow \frac{d y}{d x} = \frac{1}{x}$

Page No 10.123:

Question 26:

If f (x) is an even function, then write whether f' (x) is even or odd.

Answer:

$We have, f (x) is an even function . \Rightarrow f (- x) = f (x)$

$\Rightarrow \frac{d}{d x} \{f (- x)\} = \frac{d}{d x} \{f (x)\} \Rightarrow f' (- x) \frac{d}{d x} (- x) = f' (x) \Rightarrow f' (- x) \times (- 1) = f' (x) \Rightarrow - f' (- x) = f' (x) \Rightarrow f' (- x) = - f' (x) Thus, f' (x) is an odd function .$

Page No 10.123:

Question 27:

If f (x) is an odd function, then write whether f' (x) is even or odd.

Answer:

$We have, f (x) is an odd function . \Rightarrow f (- x) = - f (x)$

$\Rightarrow \frac{d}{d x} \{f (- x)\} = - \frac{d}{d x} \{f (x)\} \Rightarrow f' (- x) \frac{d}{d x} (- x) = - f' (x) \Rightarrow f' (- x) \times (- 1) = - f' (x) \Rightarrow - f' (- x) = - f' (x) \Rightarrow f' (- x) = f' (x) Thus, f' (x) is an even function .$

Page No 10.123:

Question 28:

If $x = 3 \sin t - \sin 3 t, y = 3 \cos t - \cos 3 t find \frac{d y}{d x} at t = \frac{π}{3}$

Answer:

$x = 3 \sin t - \sin 3 t and y = 3 \cos t - \cos 3 t \Rightarrow \frac{d x}{d t} = 3 \cos t - 3 \cos 3 t and \frac{d y}{d t} = - 3 \sin t + 3 \sin 3 t ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- 3 \sin t + 3 \sin 3 t}{3 \cos t - 3 \cos 3 t} \Rightarrow {(\frac{d y}{d x})}_{t = \frac{π}{3}} = \frac{- 3 \sin \frac{π}{3} + 3 sinπ}{3 \cos \frac{π}{3} - 3 cosπ} = \frac{- 3 \times \frac{\sqrt{3}}{2} + 0}{3 \times \frac{1}{2} + 3} = \frac{\frac{- 3 \sqrt{3}}{2}}{\frac{9}{2}} = - \frac{1}{\sqrt{3}}$

Page No 10.123:

Question 29:

If y = log (cos e^x), then find $\frac{d y}{d x} .$

Answer:

$y = \log (\cos e^{x})$

Differentiating both sides with respect to x, we get

$\frac{d y}{d x} = \frac{d}{d x} \log (\cos e^{x})$

$\Rightarrow \frac{d y}{d x} = \frac{1}{\cos e^{x}} \times \frac{d}{d x} \cos e^{x}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{\cos e^{x}} \times (- \sin e^{x}) \times \frac{d}{d x} e^{x}$

$\Rightarrow \frac{d y}{d x} = - \frac{\sin e^{x}}{\cos e^{x}} \times e^{x}$

$\Rightarrow \frac{d y}{d x} = - e^{x} \tan e^{x}$

Page No 10.123:

Question 30:

If f(x) = x + 7 and g(x) = x – 7, x ∈ R, then find $\frac{d}{d x} (f o g) (x) .$

Answer:

The given functions are f(x) = x + 7 and g(x) = x – 7, x ∈ R.

$f o g (x)$

$= f (g (x))$

$= g (x) + 7 [f (x) = x + 7]$

$= x - 7 + 7 [g (x) = x - 7]$

$= x$

$∴ \frac{d}{d x} (f o g) (x) = \frac{d}{d x} (x)$

$\Rightarrow \frac{d}{d x} (f o g) (x) = 1$

Thus, the value of $\frac{d}{d x} (f o g) (x)$ is 1.

Page No 10.17:

Question 1:

Differentiate the following functions from first principles:

e^−x

Answer:

$Let f (x) = e^{- x} \Rightarrow f (x + h) = e^{- (x + h)} \frac{d}{d x} \{f (x)\} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{e^{- (x + h)} - e^{- x}}{h} = \lim_{h \to 0} \frac{e^{- x} \times e^{- h} - e^{- x}}{h} = \lim_{h \to 0} e^{- x} \{\frac{(e^{- h} - 1)}{- h}\} \times (- 1) = - e^{- x} \lim_{h \to 0} \{\frac{(e^{- h} - 1)}{- h}\} = - e^{- x} [∵ \lim_{h \to 0} \frac{e^{- h} - 1}{- h} = 1] So, \frac{d}{d x} (e^{- x}) = - e^{- x}$

Page No 10.17:

Question 2:

Differentiate the following functions from first principles:

e^3x

Answer:

$Let f (x) = e^{3 x} \Rightarrow f (x + h) = e^{3 (x + h)} \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{e^{3 (x + h)} - e^{3 x}}{h} = \lim_{h \to 0} \frac{e^{3 x} e^{3 h} - e^{3 x}}{h} = \lim_{h \to 0} e^{3 x} \{\frac{(e^{3 h} - 1)}{3 h}\} \times 3 = 3 e^{3 x} \lim_{h \to 0} \{\frac{(e^{3 h} - 1)}{3 h}\} = 3 e^{3 x} Hence, \frac{d}{d x} (e^{3 x}) = 3 e^{3 x}$

Page No 10.17:

Question 3:

Differentiate the following functions from first principles:

e^ax^+b

Answer:

$Let f (x) = e^{a x + b} \Rightarrow f (x + h) = e^{a (x + h) + b} ∴ \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{e^{a (x + h) + b} - e^{(a x + b)}}{h} = \lim_{h \to 0} \frac{e^{a x + b} e^{a h} - e^{a x + b}}{h} = \lim_{h \to 0} e^{a x + b} \{\frac{(e^{a h} - 1)}{a h}\} \times a = a e^{a x + b} \lim_{h \to 0} \{\frac{(e^{a h} - 1)}{a h}\} = a e^{a x + b} So, \frac{d}{d x} (e^{a x + b}) = a e^{a x + b}$

Page No 10.17:

Question 4:

Differentiate the following functions from first principles:

e^cos^x

Answer:

$Let f (x) = e^{\cos x} \Rightarrow f (x + h) = e^{\cos (x + h)} ∴ \frac{d}{d x} \{f (x)\} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{e^{\cos (x + h)} - e^{\cos x}}{h} = \lim_{h \to 0} e^{\cos x} [\frac{e^{\cos (x + h) - \cos x} - 1}{h}] = \lim_{h \to 0} e^{\cos x} [\frac{e^{\cos (x + h) - \cos x} - 1}{\cos (x + h) \cos x}] \times \frac{\cos (x + h) - \cos x}{h} = \underset{h \to 0}{e^{\cos x} \lim} (\frac{\cos (x + h) - \cos x}{h}) \times \lim_{h \to 0} [\frac{e^{\cos (x + h) - \cos x} - 1}{\cos (x + h) - \cos x}] = e^{\cos x} \lim_{h \to 0} (\frac{\cos (x + h) - \cos x}{h}) [∵ \lim_{h \to 0} \frac{e^{x} - 1}{x} = 1] = e^{\cos x} \lim_{h \to 0} \{\frac{- 2 \sin (\frac{x + h + x}{2}) \sin (\frac{x + h - x}{2})}{h}\} [∵ \cos A - \cos B = - 2 \sin (\frac{A + B}{2}) \sin (\frac{A - B}{2})] = e^{\cos x} \lim_{h \to 0} \frac{- \sin (\frac{2 x + h}{2})}{1} \times \frac{\sin (\frac{h}{2})}{\frac{h}{2}} = e^{\cos x} \lim_{h \to 0} \frac{- \sin (\frac{2 x + h}{2})}{1} \underset{h \to 0}{\times \lim} \frac{\sin (\frac{h}{2})}{\frac{h}{2}} = e^{\cos x} \lim_{h \to 0} - \sin (\frac{2 x + h}{2}) [∵ \frac{\sin x}{x} = 1] = e^{\cos x} (- \sin x) = - \sin x e^{\cos x} Hence, \frac{d}{d x} (e^{\cos x}) = - \sin x e^{\cos x}$

Page No 10.17:

Question 5:

Differentiate the following functions from first principles:

$e^{\sqrt{2 x}}$

Answer:

$Let f (x) = e^{\sqrt{2 x}} \Rightarrow f (x + h) = e^{\sqrt{2 (x + h)}} ∴ \frac{d}{d x} \{f (x)\} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{e^{^{\sqrt{2 (x + h)}}} - e^{^{\sqrt{2 x}}}}{h} = \lim_{h \to 0} e^{\sqrt{2 x}} [\frac{e^{\sqrt{2 (x + h)} - \sqrt{2 x}} - 1}{h}] = e^{\sqrt{2 x}} \lim_{h \to 0} [\frac{e^{\sqrt{2 (x + h)} - \sqrt{2 x}} - 1}{\sqrt{2 (x + h)} - \sqrt{2 x}}] \times \lim_{h \to 0} \frac{\sqrt{2 (x + h)} - \sqrt{2 x}}{h} = e^{\sqrt{2 x}} \lim_{h \to 0} \frac{\sqrt{2 (x + h)} - \sqrt{2 x}}{h} [∵ \lim_{h \to 0} \frac{e^{h} - 1}{h} = 1] = e^{\sqrt{2 x}} \lim_{h \to 0} \frac{\sqrt{2 (x + h)} - \sqrt{2 x}}{h} \times \frac{\sqrt{2 (x + h)} + \sqrt{2 x}}{\sqrt{2 (x + h)} + \sqrt{2 x}} [Rationalising the numerator] = e^{\sqrt{2 x}} \lim_{h \to 0} \frac{2 (x + h) - 2 x}{h (\sqrt{2 (x + h)} + \sqrt{2 x})} = e^{\sqrt{2 x}} \lim_{h \to 0} \frac{2 x + 2 h - 2 x}{h (\sqrt{2 (x + h)} + \sqrt{2 x})} = e^{\sqrt{2 x}} \lim_{h \to 0} \frac{2 h}{h (\sqrt{2 (x + h)} + \sqrt{2 x})} = e^{\sqrt{2 x}} \lim_{h \to 0} \frac{2}{(\sqrt{2 (x + h)} + \sqrt{2 x})} = \frac{e^{\sqrt{2 x}}}{\sqrt{2 x}} Hence, \frac{d}{d x} (e^{\sqrt{2 x}}) = \frac{e^{\sqrt{2 x}}}{\sqrt{2 x}}$

Page No 10.17:

Question 6:

Differentiate the following functions from first principles:

log cos x

Answer:

$Let f (x) = \log \cos x \Rightarrow f (x + h) = \log \cos (x + h) ∴ \frac{d}{d x} \{f (x)\} = \lim_{h \to 0} \frac{f (x + h) = f (x)}{h} = \lim_{h \to 0} \frac{\log \cos (x + h) - \log \cos x}{h} = \lim_{h \to 0} \frac{\log \frac{\cos (x + h)}{\cos x}}{h} [∵ \log A - \log B = \log (\frac{A}{B})] = \lim_{h \to 0} \frac{\log [1 + \{\frac{\cos (x + h)}{\cos x} - 1\}]}{h} = \lim_{h \to 0} \frac{\log \{1 + \frac{\cos (x + h) - \cos x}{\cos x}\}}{\{\frac{\cos (x + h) - \cos x}{\cos x}\}} \times \lim_{h \to 0} \{\frac{\cos (x + h) - \cos x}{\cos x}\} = 1 \times \lim_{h \to 0} \frac{\cos (x + h) - \cos x}{\cos x \times h} [∵ \lim_{x \to 0} \frac{\log (1 + x)}{x} = 1] = \lim_{h \to 0} \frac{- 2 \sin (\frac{x + h + x}{2}) \sin (\frac{x + h - x}{2})}{\cos x \times h} = - 2 \lim_{h \to 0} \frac{\sin (\frac{2 x + h}{2}) \times \sin (\frac{h}{2})}{2 \cos x \times (\frac{h}{2})} = \frac{- 2 \sin x}{2 \cos x} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] = - \tan x So, \frac{d}{d x} (\log \cos x) = - \tan x$

Page No 10.17:

Question 7:

Differentiate the following function from first principles:

$e^{\sqrt{\cot x}}$

Answer:

$Let f (x) = e^{\sqrt{\cot x}} \Rightarrow f (x + h) = e^{\sqrt{\cot (x + h)}} ∴ \frac{d}{d x} \{f (x)\} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{e^{\sqrt{\cot (x + h)}} - e^{\sqrt{\cot x}}}{h} = \lim_{h \to 0} \frac{e^{\sqrt{\cot x}} (e^{\sqrt{\cot (x + h)} - \sqrt{\cot x}} - 1)}{h} = e^{\sqrt{\cot x}} \lim_{h \to 0} (\frac{e^{\sqrt{\cot (x + h)} - \sqrt{\cot x}} - 1}{\sqrt{\cot (x + h)} - \sqrt{\cot x}}) \times (\frac{\sqrt{\cot (x + h)} - \sqrt{\cot x}}{h}) = e^{\sqrt{\cot x}} \lim_{h \to 0} \frac{(\sqrt{\cot (x + h)} - \sqrt{\cot x})}{h} \times \frac{\sqrt{\cot (x + h)} + \sqrt{\cot x}}{\sqrt{\cot (x + h)} + \sqrt{\cot x}} [∵ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1 and rationalizing the numerator] = e^{\sqrt{\cot x}} \lim_{h \to 0} \frac{\cot (x + h) - \cot x}{h (\sqrt{\cot (x + h)} + \sqrt{\cot x})} = e^{\sqrt{\cot x}} \lim_{h \to 0} \frac{\frac{\cot (x + h) \cot x + 1}{\cot (x - x - h)}}{h (\sqrt{\cot (x + h)} + \sqrt{\cot x})} [∵ \cot (A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}] = e^{\sqrt{\cot x}} \lim_{h \to 0} \frac{\cot (x + h) \cot x + 1}{\cot (- h) \times h (\sqrt{\cot (x + h)} + \sqrt{\cot x})} = - e^{\sqrt{\cot x}} \lim_{h \to 0} \frac{\cot (x + h) \cot x + 1}{(\frac{h}{\tan h}) (\sqrt{\cot (x + h)} + \sqrt{\cot x})} = \frac{e^{\sqrt{\cot x}} \times (\cot^{2} x + 1)}{2 \sqrt{\cot x}} [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1] = - \frac{e^{\sqrt{\cot x}} \times {cosec}^{2} x}{2 \sqrt{\cot x}} [∵ (1 + \cot^{2} x) = {cosec}^{2} x] ∴ \frac{d}{d x} (e^{\sqrt{c o t x}}) = - \frac{e^{\sqrt{\cot x}} \times {cosec}^{2} x}{2 \sqrt{\cot x}}$

Page No 10.17:

Question 8:

Differentiate the following functions from first principles:

x²e^x

Answer:

$Let f (x) = x^{2} e^{x} \Rightarrow f (x + h) = {(x + h)}^{2} e^{(x + h)} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{{(x + h)}^{2} e^{(x + h)} - x^{2} e^{x}}{h} = \lim_{h \to 0} (\frac{x^{2} e^{x + h} - x^{2} e^{x}}{h} + \frac{2 x h e^{(x + h)}}{h} + \frac{h^{2} e^{(x + h)}}{h}) = \lim_{h \to 0} (\frac{x^{2} e^{x} (e^{(x + h) - x} - 1)}{h} + 2 x e^{(x + h)} + h e^{(x + h)}) = \lim_{h \to 0} (x^{2} e^{x} \frac{(e^{h} - 1)}{h} + 2 x e^{(x + h)} + h e (x + h)) = x^{2} e^{x} + 2 x e^{x} + 0 x e^{x} [∵ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1] ∴ \frac{d}{d x} (x^{2} e^{x}) = e^{x} (x^{2} + 2 x)$

Page No 10.17:

Question 9:

Differentiate the following functions from first principles:

log cosec x

Answer:

$Let f (x) = \log cosec x \Rightarrow f (x + h) = \log cosec (x + h) ∴ \frac{d}{d x} \{f (x)\} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\log cosec (x + h) - \log cosec x}{h} = \lim_{h \to 0} \frac{\log \{\frac{cosec (x + h)}{cosec x}\}}{h} = \lim_{h \to 0} \frac{\log \{1 + (\frac{\sin x}{\sin (x + h)} - 1)\}}{h} = \lim_{h \to 0} \{\frac{\log \{1 + (\frac{\sin x - \sin (x + h)}{\sin (x + h)})\}}{\{\frac{\sin x - \sin (x + h)}{\sin (x + h)}\}}\} \frac{\{\frac{\sin x - \sin (x + h)}{\sin (x + h)}\}}{h} = \lim_{h \to 0} \frac{2 \cos (\frac{x + x + h}{2}) \sin (\frac{x - x - h}{2})}{\sin (x + h) h} [∵ \lim_{x \to 0} \frac{\log (1 + x)}{x} = 1 and \sin A - \sin B = 2 \cos (\frac{A + B}{2}) \sin (\frac{A - B}{2})] = \lim_{h \to 0} \frac{2 \cos (\frac{2 x + h}{2})}{\sin (x + h) (- 2)} \{\frac{\sin (- \frac{h}{2})}{- \frac{h}{2}}\} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] = - \cot x ∴ \frac{d}{d x} (\log cosec x) = - \cot x$

Page No 10.17:

Question 10:

Differentiate the following functions from first principles:

sin⁻¹ (2x + 3)

Answer:

$Let f (x) = \sin^{- 1} (2 x + 3) \Rightarrow f (x + h) = \sin^{- 1} (2 (x + h) + 3) \Rightarrow f (x + h) = \sin^{- 1} (2 x + 2 h + 3) ∴ \frac{d}{d x} \{f (x)\} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sin^{- 1} (2 x + 2 h + 3) - \sin^{- 1} (2 x + 3)}{h} = \lim_{h \to 0} \frac{\sin^{- 1} [(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} - (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}]}{h} [∵ \sin^{- 1} x - \sin^{- 1} y = \sin^{- 1} [x \sqrt{1 - y^{2}} - y \sqrt{1 - x^{2}}]] = \lim_{h \to 0} \frac{\sin^{- 1} z}{z} \times \frac{z}{h} where, z = (2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} - (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}} and \lim_{h \to 0} \frac{\sin^{- 1} h}{h} = 1 = \lim_{h \to 0} \frac{z}{h} = \lim_{h \to 0} \frac{(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} - (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}}{h} = \lim_{h \to 0} \frac{{(2 x + 2 h + 3)}^{2} \{1 - {(2 x + 3)}^{2}\} - {(2 x + 3)}^{2} \{1 - {(2 x + 2 h + 3)}^{2}\}}{h \{(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} + (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}\}} [Rationalizing numerator] = \lim_{h \to 0} \frac{[{(2 x + 3)}^{2} + 4 h^{2} + 4 h (2 x + 3)] \{1 - {(2 x + 3)}^{2}\} - {(2 x + 3)}^{2} [1 - {(2 x + 3)}^{2} - 4 h^{2} - 4 h (2 x + 3)]}{h \{(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} + (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}\}} = \lim_{h \to 0} \frac{[{(2 x + 3)}^{2} + 4 h^{2} + 4 h (2 x + 3) - {(2 x + 3)}^{4} - 4 h^{2} {(2 x + 3)}^{2} - 4 h {(2 x + 3)}^{3} - {(2 x + 3)}^{2} + {(2 x + 3)}^{4} + 4 h^{2} {(2 x + 3)}^{2} + 4 h {(2 x + 3)}^{3}]}{h \{(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} + (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}\}} = \lim_{h \to 0} \frac{4 h [h + (2 x + 3)]}{h \{(2 x + 2 h + 3) \sqrt{1 - {(2 x + 3)}^{2}} + (2 x + 3) \sqrt{1 - {(2 x + 2 h + 3)}^{2}}\}} = \frac{4 (2 x + 3)}{(2 x + 3) \sqrt{1 - {(2 x + 3)}^{2}} + (2 x + 3) \sqrt{1 - {(2 x + 3)}^{2}}} = \frac{4 (2 x + 3)}{2 (2 x + 3) \sqrt{1 - {(2 x + 3)}^{2}}} = \frac{2}{\sqrt{1 - {(2 x + 3)}^{2}}} ∴ \frac{d}{d x} \{\sin^{- 1} (2 x + 3)\} = \frac{2}{\sqrt{1 - {(2 x + 3)}^{2}}}$

Page No 10.37:

Question 1:

Differentiate

sin (3x + 5)

Answer:

$Let y = \sin (3 x + 5) Differentiating y with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} \sin (3 x + 5) = \cos (3 x + 5) \frac{d}{d x} (3 x + 5) [using chain rule] = \cos (3 x + 5) \times 3 = 3 \cos (3 x + 5) So, \frac{d}{d x} \{\sin (3 x + 5)\} = 3 \cos (3 x + 5)$

Page No 10.37:

Question 2:

Differentiate

tan² x

Answer:

$Let y = \tan^{2} x Differentiating with respect to x we get, \frac{d y}{d x} = 2 \tan x \frac{d}{d x} (\tan x) [using chain rule] = 2 \tan x \times \sec^{2} x So, \frac{d}{d x} (\tan^{2} x) = 2 \tan x \sec^{2} x$

Page No 10.37:

Question 3:

Differentiate

tan (x° + 45°)

Answer:

$Let, y = \tan (x ° + 45 °) \Rightarrow y = \tan \{(x + 45) \frac{π}{180}\} Differentiating it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} \tan \{(x + 45) \frac{π}{180}\} = \sec^{2} \{(x + 45) \frac{π}{180}\} \times \frac{d}{d x} (x + 45) \frac{π}{180} [Using chain rule] = \frac{π}{180} \sec^{2} (x ° + 45 °) S o, \frac{d}{d x} \{\tan (x ° + 45 °)\} = \frac{π}{180} \sec^{2} (x ° + 45 °)$

Page No 10.37:

Question 4:

Differentiate

sin (log x)

Answer:

$Let y = \sin (\log x) Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} \sin (\log x) = \cos (\log x) \frac{d}{d x} (\log x) [using chain rule] = \frac{1}{x} \cos (\log x) So, \frac{d}{d x} \{\sin (\log x)\} = \frac{1}{x} \cos (\log x)$

Page No 10.37:

Question 5:

Differentiate

$e^{\sin \sqrt{x}}$

Answer:

$Let y = e^{\sin \sqrt{x}} Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} (e^{\sin \sqrt{x}}) = e^{\sin \sqrt{x}} \frac{d}{d x} (\sin \sqrt{x}) [using chain rule] = e^{\sin \sqrt{x}} \times \cos \sqrt{x} \frac{d}{d x} \sqrt{x} [using chain rule] = e^{\sin \sqrt{x}} \times \cos \sqrt{x} \times \frac{1}{2 \sqrt{x}} = \frac{\cos \sqrt{x} e^{\sin \sqrt{x}}}{2 \sqrt{x}} So, \frac{d}{d x} (e^{\sin \sqrt{x}}) = \frac{\cos \sqrt{x} e^{\sin \sqrt{x}}}{2 \sqrt{x}}$

Page No 10.37:

Question 6:

Differentiate

e^tan^x

Answer:

$Let y = e^{\tan x} Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} (e^{\tan x}) = e^{\tan x} \frac{d}{d x} (\tan x) [using chain rule] = e^{\tan x} \times \sec^{2} x So, \frac{d}{d x} (e^{\tan x}) = \sec^{2} x e^{\tan x}$

Page No 10.37:

Question 7:

Differentiate

sin² (2x + 1)

Answer:

$Let y = \sin^{2} (2 x + 1) Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} [\sin^{2} (2 x + 1)] = 2 \sin (2 x + 1) \frac{d}{d x} \sin (2 x + 1) [using chain rule] = 2 \sin (2 x + 1) \cos (2 x + 1) \frac{d}{d x} (2 x + 1) [using chain rule] = 4 \sin (2 x + 1) \cos (2 x + 1) = 2 \sin 2 (2 x + 1) [∵ \sin 2 A = 2 \sin A \cos A] = 2 \sin (4 x + 2) So, \frac{d}{d x} \{\sin^{2} (2 x + 1)\} = 2 \sin (4 x + 2)$

Page No 10.37:

Question 8:

Differentiate

log₇ (2x − 3)

Answer:

$Let y = \log_{7} (2 x - 3) \Rightarrow y = \frac{\log (2 x - 3)}{\log 7} [∵ \log_{a} b = \frac{\log b}{\log a}] Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{1}{\log 7} \frac{d}{d x} \{\log (2 x - 3)\} = \frac{1}{\log 7} \times \frac{1}{(2 x - 3)} \frac{d}{d x} (2 x - 3) [using chain rule] = \frac{2}{(2 x - 3) \log 7} Hence, \frac{d}{d x} \{\log_{7} (2 x - 3)\} = \frac{2}{(2 x - 3) \log 7}$

Page No 10.37:

Question 9:

Differentiate

tan 5x°

Answer:

$Let y = \tan 5 x ° \Rightarrow y = \tan (5 x \times \frac{π}{180}) Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} \tan (5 x \times \frac{π}{180}) = \sec^{2} (5 x \times \frac{π}{180}) \frac{d}{d x} (5 x \times \frac{π}{180}) [using chain rule] = (\frac{5 π}{180}) \sec^{2} (5 x \times \frac{π}{180}) = \frac{5 π}{180} \sec^{2} (5 x °) Hence, \frac{d}{d x} (\tan 5 x °) = \frac{5 π}{180} \sec^{2} (5 x °)$

Page No 10.37:

Question 10:

Differentiate

${2^{x}}^{3}$

Answer:

$Let y = 2^{x^{3}} Differentiate it with respect to x w e g e t, \frac{d y}{d x} = \frac{d}{d x} (2^{x^{3}}) = 2^{x^{3}} \times \log_{e} 2 \frac{d}{d x} (x^{3}) [using chain rule] = 3 x^{2} \times 2^{x^{3}} \times \log_{e} 2 Hence, \frac{d}{d x} (2^{x^{3}}) = 3 x^{2} \times 2^{x^{3}} \log_{e} 2$

Page No 10.37:

Question 11:

Differentiate

$3^{e^{x}}$

Answer:

$Let y = 3^{e^{x}} Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} (3^{e^{x}}) = 3^{e^{x}} \log 3 \frac{d}{d x} (e^{x}) [using chain rule] = e^{x} \times 3^{e^{x}} \log 3 So, \frac{d}{d x} (3^{e^{x}}) = e^{x} \times 3^{e^{x}} \log 3$

Page No 10.37:

Question 12:

Differentiate

log_x 3

Answer:

$Let y = \log_{x} 3 \Rightarrow y = \frac{\log 3}{\log x} [∵ \log_{a} b = \frac{\log b}{\log a}] Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} (\frac{\log 3}{\log x}) = \log 3 \frac{d}{d x} {(\log x)}^{- 1} = \log 3 \times [- 1 {(\log x)}^{- 2}] \frac{d}{d x} (\log x) [using chain rule] = - \frac{\log 3}{{(\log x)}^{2}} \times \frac{1}{x} = - {(\frac{\log 3}{\log x})}^{2} \times \frac{1}{x} \times \frac{1}{\log 3} = - \frac{1}{x \log 3 {(\log_{3} x)}^{2}} [∵ \frac{\log b}{\log a} = \log_{a} b] So, \frac{d}{d x} (\log_{x} 3) = - \frac{1}{x \log 3 {(\log_{3} x)}^{2}}$

Page No 10.37:

Question 13:

Differentiate

$3^{x^{2} + 2 x}$

Answer:

$Let y = 3^{x^{2} + 2 x} Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} (3^{x^{2} + 2 x}) = 3^{x^{2} + 2 x} \times \log_{e} 3 \frac{d}{d x} (x^{2} + 2 x) [using chain rule] = (2 x + 2) 3^{x^{2} + 2 x} \log_{e} 3 So, \frac{d}{d x} (3^{x^{2} + 2 x}) = (2 x + 2) 3^{x^{2} + 2 x} \log_{e} 3$

Page No 10.37:

Question 14:

Differentiate

$\sqrt{\frac{a^{2} - x^{2}}{a^{2} + x^{2}}}$

Answer:

$Let y = \sqrt{\frac{a^{2} - x^{2}}{a^{2} + x^{2}}} \Rightarrow y = {(\frac{a^{2} - x^{2}}{a^{2} + x^{2}})}^{\frac{1}{2}} Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} {(\frac{a^{2} - x^{2}}{a^{2} + x^{2}})}^{\frac{1}{2}} = \frac{1}{2} {(\frac{a^{2} - x^{2}}{a^{2} + x^{2}})}^{\frac{1}{2} - 1} \times \frac{d}{d x} (\frac{a^{2} - x^{2}}{a^{2} + x^{2}}) [Using chain rule] = \frac{1}{2} {(\frac{a^{2} - x^{2}}{a^{2} + x^{2}})}^{\frac{- 1}{2}} \times \{\frac{(a^{2} + x^{2}) \frac{d}{d x} (a^{2} - x^{2}) - (a^{2} - x^{2}) \frac{d}{d x} (a^{2} + x^{2})}{{(a^{2} + x^{2})}^{2}}\} = \frac{1}{2} {(\frac{a^{2} + x^{2}}{a^{2} - x^{2}})}^{\frac{1}{2}} \{\frac{- 2 x (a^{2} + x^{2}) - 2 x (a^{2} - x^{2})}{{(a^{2} + x^{2})}^{2}}\} = \frac{1}{2} {(\frac{a^{2} + x^{2}}{a^{2} - x^{2}})}^{\frac{1}{2}} \{\frac{- 2 x a^{2} - 2 x^{3} - 2 x a^{2} + 2 x^{3}}{{(a^{2} + x^{2})}^{2}}\} = \frac{1}{2} {(\frac{a^{2} + x^{2}}{a^{2} - x^{2}})}^{\frac{1}{2}} \{\frac{- 4 x a^{2}}{{(a^{2} + x^{2})}^{2}}\} = \frac{- 2 x a^{2}}{\sqrt{a^{2} - x^{2}} {(a^{2} + x^{2})}^{\frac{3}{2}}} So, \frac{d}{d x} (\sqrt{\frac{a^{2} - x^{2}}{a^{2} + x^{2}}}) = \frac{- 2 a^{2} x}{\sqrt{a^{2} - x^{2}} {(a^{2} + x^{2})}^{\frac{3}{2}}}$

Page No 10.37:

Question 15:

Differentiate

$3^{x \log x}$

Answer:

$Let y = 3^{x \log x} Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} (3^{x \log x}) = 3^{x \log x} \times \log_{e} 3 \frac{d}{d x} (x \log x) [Using chain rule] = 3^{x \log x} \times \log_{e} 3 [x \frac{d}{d x} (\log x) + \log x \frac{d}{d x} (x)] = 3^{x \log x} \times \log_{e} 3 [\frac{x}{x} + \log x] = 3^{x \log x} (1 + \log x) \times \log_{e} 3 So, \frac{d}{d x} (3^{x \log x}) = 3^{x \log x} (1 + \log x) \log_{e} 3$

Page No 10.37:

Question 16:

Differentiate

$\sqrt{\frac{1 + \sin x}{1 - \sin x}}$

Answer:

$Let y = \sqrt{\frac{1 + \sin x}{1 - \sin x}} Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} {(\frac{1 + \sin x}{1 - \sin x})}^{\frac{1}{2}} = \frac{1}{2} {(\frac{1 + \sin x}{1 - \sin x})}^{\frac{1}{2} - 1} \frac{d}{d x} (\frac{1 + \sin x}{1 - \sin x}) = \frac{1}{2} {(\frac{1 - \sin x}{1 + \sin x})}^{\frac{1}{2}} [\frac{(1 - \sin x) (\cos x) - (1 + \sin x) (- \cos x)}{{(1 - \sin x)}^{2}}] = \frac{1}{2} \frac{{(1 - \sin x)}^{\frac{1}{2}}}{{(1 + \sin x)}^{\frac{1}{2}}} [\frac{\cos x - \cos x \sin x + \cos x + \sin x \cos x}{{(1 - \sin x)}^{2}}] = \frac{1}{2} \times \frac{2 \cos x}{\sqrt{1 + \sin x} (1 - \sin x) \frac{3}{2}} = \frac{\cos x}{\sqrt{1 + \sin x} (1 - \sin x) \frac{3}{2}} = \frac{\cos x}{\sqrt{1 + \sin x} \sqrt{1 - \sin x} (1 - \sin x)} = \frac{\cos x}{\sqrt{1 - \sin^{2} x} \times (1 - \sin x)} = \frac{\cos x}{\cos x (1 - \sin x)} [Using 1 - \sin^{2} x = \cos^{2} x] = \frac{1}{(1 - \sin x)} \times \frac{(1 + \sin x)}{(1 + \sin x)} = \frac{(1 + \sin x)}{(1 - \sin^{2} x)} = \frac{1 + \sin x}{\cos^{2} x} = \frac{1}{\cos x} (\frac{1}{\cos x} + \frac{\sin x}{\cos x}) = \sec x (\sec x + \tan x) Hence, \frac{d y}{d x} = \sec x (\sec x + \tan x)$

Page No 10.37:

Question 17:

Differentiate

$\sqrt{\frac{1 - x^{2}}{1 + x^{2}}}$

Answer:

$Let y = \sqrt{\frac{1 - x^{2}}{1 + x^{2}}} \Rightarrow y = {(\frac{1 - x^{2}}{1 + x^{2}})}^{\frac{1}{2}} Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} {(\frac{1 - x^{2}}{1 + x^{2}})}^{\frac{1}{2}} = \frac{1}{2} {(\frac{1 - x^{2}}{1 + x^{2}})}^{\frac{1}{2} - 1} \times \frac{d}{d x} (\frac{1 - x^{2}}{1 + x^{2}}) [Using chain rule] = \frac{1}{2} {(\frac{1 - x^{2}}{1 + x^{2}})}^{\frac{- 1}{2}} \times \{\frac{(1 + x^{2}) \frac{d}{d x} (1 - x^{2}) - (1 - x^{2}) \frac{d}{d x} (1 + x^{2})}{{(1 + x^{2})}^{2}}\} = \frac{1}{2} {(\frac{1 + x^{2}}{1 - x^{2}})}^{\frac{1}{2}} \{\frac{- 2 x (1 + x^{2}) - 2 x (1 - x^{2})}{{(1 + x^{2})}^{2}}\} = \frac{1}{2} {(\frac{1 + x^{2}}{1 - x^{2}})}^{\frac{1}{2}} \{\frac{- 2 x - 2 x^{3} - 2 x + 2 x^{3}}{{(1 + x^{2})}^{2}}\} = \frac{1}{2} {(\frac{1 + x^{2}}{1 - x^{2}})}^{\frac{1}{2}} \{\frac{- 4 x}{{(1 + x^{2})}^{2}}\} = \frac{- 2 x}{\sqrt{1 - x^{2}} {(1 + x^{2})}^{\frac{3}{2}}}$

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Question 18:

Differentiate

(log sin x)²

Answer:

$Let y = {(\log \sin x)}^{2} Differentiate with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} {(\log \sin x)}^{2} = 2 (\log \sin x) \frac{d}{d x} (\log \sin x) = 2 (\log \sin x) \times \frac{1}{\sin x} \frac{d}{d x} (\sin x) = 2 (\log \sin x) \times \frac{1}{\sin x} \times \cos x = 2 (\log \sin x) \cot x So, \frac{d}{d x} {(\log \sin x)}^{2} = 2 (\log \sin x) \cot x$

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Question 19:

Differentiate

$\sqrt{\frac{1 + x}{1 - x}}$

Answer:

$Let y = \sqrt{\frac{1 + x}{1 - x}} \Rightarrow y = {(\frac{1 + x}{1 - x})}^{\frac{1}{2}} Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} {(\frac{1 + x}{1 - x})}^{\frac{1}{2}} = \frac{1}{2} {(\frac{1 + x}{1 - x})}^{\frac{1}{2} - 1} \times \frac{d}{d x} (\frac{1 + x}{1 - x}) [Using chain rule] = \frac{1}{2} {(\frac{1 + x}{1 - x})}^{\frac{- 1}{2}} \times \{\frac{(1 - x) \frac{d}{d x} (1 + x) - (1 + x) \frac{d}{d x} (1 - x)}{{(1 - x)}^{2}}\} [Using quotient rule] = \frac{1}{2} {(\frac{1 - x}{1 + x})}^{\frac{1}{2}} \{\frac{(1 - x) (1) - (1 + x) (- 1)}{{(1 - x)}^{2}}\} = \frac{1}{2} {(\frac{1 - x}{1 + x})}^{\frac{1}{2}} \{\frac{1 - x + 1 + x}{{(1 - x)}^{2}}\} = \frac{1}{2} \frac{{(1 - x)}^{\frac{1}{2}}}{{(1 + x)}^{\frac{1}{2}}} \times \frac{2}{{(1 - x)}^{2}} = \frac{1}{\sqrt{1 + x} {(1 - x)}^{\frac{3}{2}}} So, \frac{d}{d x} (\sqrt{\frac{1 + x}{1 - x}}) = \frac{1}{\sqrt{1 + x} {(1 - x)}^{\frac{3}{2}}}$

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Question 20:

Differentiate

$\sin (\frac{1 + x^{2}}{1 - x^{2}})$

Answer:

$Let y = \sin (\frac{1 + x^{2}}{1 - x^{2}}) Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} [\sin (\frac{1 + x^{2}}{1 - x^{2}})] = \cos x (\frac{1 + x^{2}}{1 - x^{2}}) \frac{d}{d x} (\frac{1 + x^{2}}{1 - x^{2}}) [Using chain rule] = \cos x (\frac{1 + x^{2}}{1 - x^{2}}) [\frac{(1 - x^{2}) \frac{d}{d x} (1 + x^{2}) - (1 + x^{2}) \frac{d}{d x} (1 - x^{2})}{{(1 - x^{2})}^{2}}] [Using quotient rule] = \cos x (\frac{1 + x^{2}}{1 - x^{2}}) [\frac{(1 - x^{2}) (2 x) - (1 + x^{2}) (- 2 x)}{{(1 - x^{2})}^{2}}] = \cos x (\frac{1 + x^{2}}{1 - x^{2}}) [\frac{2 x - 2 x^{3} + 2 x + 2 x^{3}}{{(1 - x^{2})}^{2}}] = \frac{4 x}{{(1 - x^{2})}^{2}} \cos x (\frac{1 + x^{2}}{1 - x^{2}}) So, \frac{d}{d x} \{\sin (\frac{1 + x^{2}}{1 - x^{2}})\} = \frac{4 x}{{(1 - x^{2})}^{2}} \cos x (\frac{1 + x^{2}}{1 - x^{2}})$

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Question 21:

Differentiate

$e^{3 x} \cos 2 x$

Answer:

$Let y = e^{3 x} \cos 2 x Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} (e^{3 x} \cos 2 x) = e^{3 x} \times \frac{d}{d x} (\cos 2 x) + \cos 2 x \frac{d}{d x} (e^{3 x}) [Using product rule] = e^{3 x} \times (- \sin 2 x) \frac{d}{d x} (2 x) + \cos 2 x e^{3 x} \frac{d}{d x} (3 x) [Using chain rule] = - 2 e^{3 x} \sin 2 x + 3 e^{3 x} \cos 2 x = e^{3 x} (3 \cos 2 x - 2 \sin 2 x) So, \frac{d}{d x} (e^{3 x} \cos 2 x) = e^{3 x} (3 \cos 2 x - 2 \sin 2 x)$

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Question 22:

Differentiate

sin (log sin x)

Answer:

$Let y = \sin (\log \sin x) Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} \sin (\log \sin x) = \cos (\log \sin x) \frac{d}{d x} (\log \sin x) [Using chain rule] = \cos (\log \sin x) \times \frac{1}{\sin x} \frac{d}{d x} \sin x [Using chain rule] = \cos (\log \sin x) \frac{\cos x}{\sin x} = \cos (\log \sin x) \cot x Hence, \frac{d}{d x} \sin (\log \sin x) = \cos (\log \sin x) \cot x$

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Question 23:

Differentiate

$e^{\tan 3 x}$

Answer:

$Let y = e^{\tan 3 x} Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} (e^{\tan 3 x}) = e^{\tan 3 x} \frac{d}{d x} (\tan 3 x) = e^{\tan 3 x} \sec^{2} 3 x \times \frac{d}{d x} (3 x) = e^{\tan 3 x} \sec^{2} 3 x \times 3 So, \frac{d}{d x} (e^{\tan 3 x}) = 3 e^{\tan 3 x} \sec^{2} 3 x$

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Question 24:

Differentiate

$e^{\sqrt{\cot x}}$

Answer:

$Let y = e^{\sqrt{\cot x}} \Rightarrow y = e^{{(\cot x)}^{\frac{1}{2}}} Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} (e^{{(\cot x)}^{\frac{1}{2}}}) = e^{{(\cot x)}^{\frac{1}{2}}} \times \frac{d}{d x} {(\cot x)}^{\frac{1}{2}} [Using chain rule] = e^{\sqrt{\cot x}} \times \frac{1}{2} {(\cot x)}^{\frac{1}{2} - 1} \frac{d}{d x} (\cot x) = - \frac{e^{\sqrt{\cot x}} \times {cosec}^{2} x}{2 \sqrt{\cot x}} So, \frac{d}{d x} (e^{\sqrt{\cot x}}) = - \frac{e^{\sqrt{\cot x}} \times {cosec}^{2} x}{2 \sqrt{\cot x}}$

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Question 25:

Differentiate

$\log (\frac{\sin x}{1 + \cos x})$

Answer:

$Let y = \log (\frac{\sin x}{1 + \cos x}) Differentiate it with respect to x, we get \frac{d y}{d x} = \frac{d}{d x} \log (\frac{\sin x}{1 + \cos x}) = \frac{1}{(\frac{\sin x}{1 + \cos x})} \times \frac{d}{d x} (\frac{\sin x}{1 + \cos x}) [Using chain rule] = (\frac{1 + \cos x}{\sin x}) [\frac{(1 + \cos x) \frac{d}{d x} (\sin x) - \sin x \frac{d}{d x} (1 + \cos x)}{{(1 + \cos x)}^{2}}] [Using quotient rule] = (\frac{1 + \cos x}{\sin x}) [\frac{(1 + \cos x) (\cos x) - \sin x (- \sin x)}{{(1 + \cos x)}^{2}}] = (\frac{1 + \cos x}{\sin x}) [\frac{\cos x + \cos^{2} x + \sin^{2} x}{{(1 + \cos x)}^{2}}] = (\frac{1 + \cos x}{\sin x}) [\frac{(1 + \cos x)}{{(1 + \cos x)}^{2}}] = \frac{1}{\sin x} = cosec x So, \frac{d}{d x} \{\log (\frac{\sin x}{1 + \cos x})\} = cosec x$

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Question 26:

Differentiate

$\log \sqrt{\frac{1 - \cos x}{1 + \cos x}}$

Answer:

$Let y = \log \sqrt{\frac{1 - \cos x}{1 + \cos x}} \Rightarrow y = \log {(\frac{1 - \cos x}{1 + \cos x})}^{\frac{1}{2}} \Rightarrow y = \frac{1}{2} \log (\frac{1 - \cos x}{1 + \cos x}) [using \log a^{b} = b \log a] Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} \{\frac{1}{2} \log (\frac{1 - \cos x}{1 + \cos x})\} = \frac{1}{2} \times \frac{1}{(\frac{1 - \cos x}{1 + \cos x})} \times \frac{d}{d x} (\frac{1 - \cos x}{1 + \cos x}) [Using chain rule] = \frac{1}{2} (\frac{1 + \cos x}{1 - \cos x}) [\frac{(1 + \cos x) \frac{d}{d x} (1 - \cos x) - (1 - \cos x) \frac{d}{d x} (1 + \cos x)}{{(1 + \cos x)}^{2}}] [Using quotient rule] = \frac{1}{2} (\frac{1 + \cos x}{1 - \cos x}) [\frac{(1 + \cos x) (\sin x) - (1 - \cos x) (- \sin x)}{{(1 + \cos x)}^{2}}] = \frac{1}{2} (\frac{1 + \cos x}{1 - \cos x}) [\frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{{(1 + \cos x)}^{2}}] = \frac{1}{2} (\frac{1 + \cos x}{1 - \cos x}) [\frac{2 \sin x}{{(1 + \cos x)}^{2}}] = \frac{\sin x}{(1 - \cos x) (1 + \cos x)} = \frac{\sin x}{1 - \cos^{2} x} = \frac{\sin x}{\sin^{2} x} = \frac{1}{\sin x} = cosec x So, \frac{d}{d x} (\log \sqrt{\frac{1 - \cos x}{1 + \cos x}}) = cosec x$

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Question 27:

Differentiate

$\tan (e^{\sin x})$

Answer:

$Let y = \tan (e^{\sin x}) Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} [\tan (e^{\sin x})] = \sec^{2} (e^{\sin x}) \frac{d}{d x} (e^{\sin x}) [Using chain rule] = \sec^{2} (e^{\sin x}) \times e^{\sin x} \times \frac{d}{d x} (\sin x) = \cos x \sec^{2} (e^{\sin x}) \times e^{\sin x} So, \frac{d}{d x} \{\tan (e^{\sin x})\} = \cos x \sec^{2} (e^{\sin x}) e^{\sin x}$

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Question 28:

Differentiate

$\log (x + \sqrt{x^{2} + 1})$

Answer:

$Let y = \log (x + \sqrt{x^{2} + 1}) Differentiate with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} \log (x + \sqrt{x^{2} + 1}) = \frac{1}{x + \sqrt{x^{2} + 1}} \frac{d}{d x} (x + {(x^{2} + 1)}^{\frac{1}{2}}) [Using chain rule] = \frac{1}{x + \sqrt{x^{2} + 1}} [1 + \frac{1}{2} {(x^{2} + 1)}^{\frac{1}{2} - 1} \frac{d}{d x} (x^{2} + 1)] = \frac{1}{x + \sqrt{x^{2} + 1}} [1 + \frac{1}{2 \sqrt{x^{2} + 1}} \times 2 x] = \frac{1}{x + \sqrt{x^{2} + 1}} [\frac{\sqrt{x^{2} + 1} + x}{\sqrt{x^{2} + 1}}] = \frac{1}{\sqrt{x^{2} + 1}} So, \frac{d}{d x} \{\log (x + \sqrt{x^{2} + 1})\} = \frac{1}{\sqrt{x^{2} + 1}}$

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Question 29:

Differentiate

$\frac{e^{x} \log x}{x^{2}}$

Answer:

$Let y = \frac{e^{x} \log x}{x^{2}} Differentiate with respect to x we get, \frac{d y}{d x} = \frac{x^{2} \frac{d}{d x} (e^{x} \log x) - (e^{x} \log x) \frac{d}{d x} x^{2}}{{(x^{2})}^{2}} [Using quotient rule] = \frac{x^{2} \{e^{x} \frac{d}{d x} (\log x) + \log x \frac{d}{d x} (e^{x})\} - e^{x} \log x \times 2 x}{x^{4}} [Using product rule] = \frac{x^{2} [\frac{e^{x}}{x} + e^{x} \log x] - 2 x e^{x} \log x}{x^{4}} = \frac{\frac{x^{2} e^{x} (1 + x \log x)}{x} - 2 x e^{x} \log x}{x^{4}} = \frac{x e^{x} [1 + x \log x - 2 \log x]}{x^{4}} = \frac{x e^{x}}{x^{3}} [\frac{1}{x} + \frac{x \log x}{x} - \frac{2 \log x}{x}] = e^{x} x^{- 2} [\frac{1}{x} + \log x - \frac{2}{x} \log x] So, \frac{d}{d x} [\frac{e^{x} \log x}{x^{2}}] = e^{x} x^{- 2} [\frac{1}{x} + \log x - \frac{2}{x} \log x]$

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Question 30:

Differentiate

$\log (cosec x - \cot x)$

Answer:

$Let y = \log (cosec x - \cot x) Differentiate it with respect to x we get, \frac{d y}{d x} = \frac{d}{d x} \log (cosec x - \cot x) = \frac{1}{(cosec x - \cot x)} \frac{d}{d x} (cosec x - \cot x) = \frac{1}{(cosec x - \cot x)} \times (- cosec x \cot x + {cosec}^{2} x) = \frac{cosec x (cosec x - \cot x)}{(cosec x - \cot x)} = cosec x So, \frac{d}{d x} \{\log (cosec x - \cot x)\} = cosec x$

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Question 31:

Differentiate

$\frac{e^{2 x} + e^{- 2 x}}{e^{2 x} - e^{- 2 x}}$

Answer:

$Let y = \frac{e^{2 x} + e^{- 2 x}}{e^{2 x} - e^{- 2 x}}$
Differentiate with respect to x we get,
$\frac{d y}{d x} = \frac{d}{d x} [\frac{e^{2 x} + e^{- 2 x}}{e^{2 x} - e^{- 2 x}}] = [\frac{(e^{2 x} - e^{- 2 x}) \frac{d}{d x} (e^{2 x} + e^{- 2 x}) - (e^{2 x} + e^{- 2 x}) \frac{d}{d x} (e^{2 x} - e^{- 2 x})}{{(e^{2 x} - e^{- 2 x})}^{2}}] [Using quotient rule and chain rule] = \frac{(e^{2 x} - e^{- 2 x}) [e^{2 x} \frac{d}{d x} (2 x) + e^{- 2 x} \frac{d}{d x} (- 2 x)] - (e^{2 x} + e^{- 2 x}) [e^{2 x} \frac{d}{d x} (2 x) - e^{- 2 x} \frac{d}{d x} (- 2 x)]}{{(e^{2 x} - e^{- 2 x})}^{2}} = \frac{(e^{2 x} - e^{- 2 x}) (2 e^{2 x} - 2 e^{- 2 x}) - (e^{2 x} + e^{- 2 x}) (2 e^{2 x} + 2 e^{- 2 x})}{{(e^{2 x} - e^{- 2 x})}^{2}} = \frac{2 {(e^{2 x} - e^{- 2 x})}^{2} - 2 {(e^{2 x} + e^{- 2 x})}^{2}}{{(e^{2 x} - e^{- 2 x})}^{2}} = \frac{2 [e^{4 x} + e^{- 4 x} - 2 e^{2 x} e^{- 2 x} - e^{4 x} - e^{- 4 x} - 2 e^{2 x} e^{- 2 x}]}{{(e^{2 x} - e^{- 2 x})}^{2}} = \frac{- 8}{{(e^{2 x} - e^{- 2 x})}^{2}} So, \frac{d}{d x} (\frac{e^{2 x} + e^{- 2 x}}{e^{2 x} - e^{- 2 x}}) = \frac{- 8}{{(e^{2 x} - e^{- 2 x})}^{2}}$

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Question 32:

Differentiate

$\log (\frac{x^{2} + x + 1}{x^{2} - x + 1})$

Answer:

$Let y = \log (\frac{x^{2} + x + 1}{x^{2} - x + 1})$

Differentiate with respect of x we get,

$\frac{d y}{d x} = \frac{d}{d x} [\log (\frac{x^{2} + x + 1}{x^{2} - x + 1})] = \frac{1}{(\frac{x^{2} + x + 1}{x^{2} - x + 1})} \frac{d}{d x} (\frac{x^{2} + x + 1}{x^{2} - x + 1}) [Using chain rule and quotient rule] = (\frac{x^{2} - x + 1}{x^{2} + x + 1}) [\frac{(x^{2} - x + 1) \frac{d}{d x} (x^{2} + x + 1) - (x^{2} + x + 1) \frac{d}{d x} (x^{2} - x + 1)}{{(x^{2} - x + 1)}^{2}}] = (\frac{x^{2} - x + 1}{x^{2} + x + 1}) [\frac{(x^{2} - x + 1) (2 x + 1) - (x^{2} + x + 1) (2 x - 1)}{{(x^{2} - x + 1)}^{2}}] = (\frac{x^{2} - x + 1}{x^{2} + x + 1}) [\frac{2 x^{3} - 2 x^{2} + 2 x + x^{2} - x + 1 - 2 x^{3} - 2 x^{2} - 2 x + x^{2} + x + 1}{{(x^{2} - x + 1)}^{2}}] = \frac{- 4 x^{2} + 2 x^{2} + 2}{(x^{2} + x + 1) (x^{2} - x + 1)} = \frac{- 4 x^{2} + 2 x^{2} + 2}{{(x^{2} + 1)}^{2} - {(x)}^{2}} = \frac{- 2 (x^{2} - 1)}{x^{4} + 1 + 2 x^{2} - x^{2}} = \frac{- 2 (x^{2} - 1)}{x^{4} + x^{2} + 1} So, \frac{d}{d x} \{\log (\frac{x^{2} + x + 1}{x^{2} - x + 1})\} = \frac{- 2 (x^{2} - 1)}{x^{4} + x^{2} + 1}$

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Question 33:

Differentiate

$\tan^{- 1} (e^{x})$

Answer:

$Let y = \tan^{- 1} (e^{x})$
Differentiate it with respect to x we get,
$\frac{d y}{d x} = \frac{d}{d x} (\tan^{- 1} e^{x}) = \frac{1}{1 + {(e^{x})}^{2}} \frac{d}{d x} (e^{x}) [Using chain rule] = \frac{1}{1 + e^{2 x}} \times e^{x} = \frac{e^{x}}{1 + e^{2 x}} So, \frac{d}{d x} (\tan^{- 1} e^{x}) = \frac{e^{x}}{1 + e^{2 x}}$

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Question 34:

Differentiate

$e^{\sin^{- 1} 2 x}$

Answer:

$Let y = e^{\sin^{- 1} 2 x}$
Differentiate it with respect to x we get,
$\frac{d y}{d x} = \frac{d}{d x} (e^{\sin^{- 1} 2 x}) = e^{\sin^{- 1} 2 x} \times \frac{d}{d x} (\sin^{- 1} 2 x) [Using chain rule] = e^{\sin^{- 1} 2 x} \times \frac{1}{\sqrt{1 - {(2 x)}^{2}}} \frac{d}{d x} (2 x) = \frac{2 e^{\sin^{- 1} 2 x}}{\sqrt{1 - 4 x^{2}}} So, \frac{d}{d x} (e^{\sin^{- 1} 2 x}) = \frac{2 e^{\sin^{- 1} 2 x}}{\sqrt{1 - 4 x^{2}}}$

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Question 35:

Differentiate

$\sin (2 \sin^{- 1} x)$

Answer:

$Let y = \sin (2 \sin^{- 1} x)$
Differentiate it with respect to x we get,
$\frac{d y}{d x} = \frac{d}{d x} [\sin (2 \sin^{- 1} x)] = \cos (2 \sin^{- 1} x) \frac{d}{d x} (2 \sin^{- 1} x) [Using chain rule] = \cos (2 \sin^{- 1} x) \times 2 \frac{1}{\sqrt{1 - x^{2}}} = \frac{2 \cos (2 \sin^{- 1} x)}{\sqrt{1 - x^{2}}} So, \frac{d}{d x} [\sin (2 \sin^{- 1} x)] = \frac{2 \cos (2 \sin^{- 1} x)}{\sqrt{1 - x^{2}}}$

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Question 36:

Differentiate

$e^{\tan^{- 1} \sqrt{x}}$

Answer:

$Let y = e^{\tan^{- 1} \sqrt{x}}$
Differentiate it with respect to x we get,
$\frac{d y}{d x} = \frac{d}{d x} (e^{\tan^{- 1} \sqrt{x}}) = e^{\tan^{- 1} \sqrt{x}} \frac{d}{d x} (\tan^{- 1} \sqrt{x}) [Using chain rule] = e^{\tan^{- 1} \sqrt{x}} \times \frac{1}{1 + {(\sqrt{x})}^{2}} \frac{d}{d x} (\sqrt{x}) = \frac{e^{\tan^{- 1} \sqrt{x}}}{1 + x} \times \frac{1}{2 \sqrt{x}} = \frac{e^{\tan^{- 1} \sqrt{x}}}{2 \sqrt{x} (1 + x)} So, \frac{d}{d x} (e^{\tan^{- 1} \sqrt{x}}) = \frac{e^{\tan^{- 1} \sqrt{x}}}{2 \sqrt{x} (1 + x)}$

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Question 37:

Differentiate

$\sqrt{\tan^{- 1} (\frac{x}{2})}$

Answer:

$Let y = \sqrt{\tan^{- 1} (\frac{x}{2})} \Rightarrow y = {\{\tan^{- 1} (\frac{x}{2})\}}^{\frac{1}{2}}$
Differentiate it with respect to x we get,
$\frac{d y}{d x} = \frac{d}{d x} {\{\tan^{- 1} (\frac{x}{2})\}}^{\frac{1}{2}} = \frac{1}{2} {\{\tan^{- 1} (\frac{x}{2})\}}^{\frac{1}{2} - 1} \frac{d}{d x} (\tan^{- 1} \frac{x}{2}) [Using chain rule] = \frac{1}{2} {\{\tan^{- 1} (\frac{x}{2})\}}^{\frac{- 1}{2}} \times \frac{1}{1 + {(\frac{x}{2})}^{2}} \times \frac{d}{d x} (\frac{x}{2}) = \frac{4}{4 (4 + x^{2}) \sqrt{\tan^{- 1} (\frac{x}{2})}} = \frac{1}{(4 + x^{2}) \sqrt{\tan^{- 1} (\frac{x}{2})}} So, \frac{d}{d x} \{\sqrt{\tan^{- 1} (\frac{x}{2})}\} = \frac{1}{(4 + x^{2}) \sqrt{\tan^{- 1} (\frac{x}{2})}}$

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Question 38:

Differentiate

$\log (\tan^{- 1} x)$

Answer:

$Let y = \log (\tan^{- 1} x)$
Differentiate it with respect to x we get,
$\frac{d y}{d x} = \frac{d}{d x} \log (\tan^{- 1} x) = \frac{1}{\tan^{- 1} x} \times \frac{d}{d x} (\tan^{- 1} x) [Using chain rule] = \frac{1}{(1 + x^{2}) \tan^{- 1} x} So, \frac{d}{d x} \{\log (\tan^{- 1} x)\} = \frac{1}{(1 + x^{2}) \tan^{- 1} x}$

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Question 39:

Differentiate

$\frac{2^{x} \cos x}{{(x^{2} + 3)}^{2}}$

Answer:

$Let y = \frac{2^{x} \cos x}{{(x^{2} + 3)}^{2}}$
Differentiate it with respect to x we get,
$\frac{d y}{d x} = \frac{d}{d x} [\frac{2^{x} \cos x}{{(x^{2} + 3)}^{2}}] = [\frac{{(x^{2} + 3)}^{2} \frac{d}{d x} (2^{x} \cos x) - (2^{x} \cos x) \frac{d}{d x} {(x^{2} + 3)}^{2}}{{[{(x^{2} + 3)}^{2}]}^{2}}] [Using quotient rule] = [()]$