Differentiate the following by using first principle -

i). tan (2x + 1)

ii). √ tan x

iii). x²sin x

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Question

Differentiate the following by using first principle -
i) tan (2x + 1)
ii) √ tan x
iii) x2 sin x

Santosh Kumar · Accepted Answer

1 The given function is f(x) = tan (2x + 1) f (x) = tan (2x + 1) âˆ´ f (x+ h) = tan [2 (x + h) + 1] $\frac{d}{d x} [f (x)] = f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\tan [2 (x + h) + 1] - \tan (2 x + 1)}{h} = \lim_{h \to 0} \frac{\tan [2 (x + h) + 1 - 2 x - 1] {1 + \tan [2 (x + h) + 1] \cdot \tan (2 x + 1)}}{h} = \lim_{h \to 0} \frac{(\tan 2 h) {1 + \tan [2 (x + h) + 1] \cdot \tan (2 x + 1)}}{h} = \lim_{h \to 0} 2 (\frac{\tan 2 h}{2 h}) \cdot \lim_{h \to 0} {1 + \tan [2 (x + h) + 1] \cdot \tan (2 x + 1)} = 2 [1 + \tan (2 x + 1) \cdot \tan (2 x + 1)] = 2 [1 + \tan^{2} (2 x + 1)] = 2 \sec^{2} (2 x + 1)$ 2 The given function is $f (x) = \sqrt{\tan x}$ $f (x) = \sqrt{\tan x} ∴ f (x + h) = \sqrt{\tan (x + h)} \frac{d}{d x} [f (x)] = f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{\sqrt{\tan (x + h)} - \sqrt{\tan x}}{h} = \lim_{h \to 0} \frac{\sqrt{\tan (x + h)} - \sqrt{\tan x}}{h} \times \frac{\sqrt{\tan (x + h)} + \sqrt{\tan x}}{\sqrt{\tan (x + h)} + \sqrt{\tan x}} = \lim_{h \to 0} \frac{(\sqrt{\tan (x + h)})^{2} - (\sqrt{\tan x})^{2}}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x})} = \lim_{h \to 0} \frac{\tan (x + h) - \tan x}{h (\sqrt{\tan (x + h)} + \sqrt{\tan x})} = \lim_{h \to 0} \frac{\tan (x + h) - \tan x}{h} \times \lim_{h \to 0} \frac{1}{(\sqrt{\tan (x + h)} + \sqrt{\tan x})} = \frac{1}{(\sqrt{\tan x} + \sqrt{\tan x})} \lim_{h \to 0} \frac{1}{h} (\frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x}) = \frac{1}{2 \sqrt{\tan x}} \lim_{h \to 0} \frac{1}{h} [\frac{\sin (x + h) \cos x - \cos (x + h) \sin x}{\cos x \cos (x + h)}] = \frac{1}{2 \sqrt{\tan x}} \lim_{h \to 0} \frac{\sin (x + h - x)}{h \cos x \cos (x + h)} [∵ \sin (A - B) = \sin A \cos B - \cos A \sin B] = \frac{1}{2 \sqrt{\tan x}} \lim_{h \to 0} \frac{\sin h}{h} \times \lim_{h \to 0} \frac{1}{\cos x \cos (x + h)} = \frac{1}{2 \sqrt{\tan x}} \times 1 \times \frac{1}{\cos^{2} x} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] = \frac{\sec^{2} x}{2 \sqrt{\tan x}}$ 3 The given function is $f (x) = x^{2} \sin x$ $f (x) = x^{2} \sin x ∴ f (x + h) = {(x + h)}^{2} \sin (x + h) \frac{d}{d x} [f (x)] = f^{'} (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{(x + h)^{2} \sin (x + h) - x^{2} \sin x}{h} = \lim_{h \to 0} \frac{(x^{2} + 2 h x + h^{2}) \sin (x + h) - x^{2} \sin x}{h} = \lim_{h \to 0} \frac{[x^{2} \sin (x + h) - x^{2} \sin x] + 2 h x \sin (x + h) + h^{2} \sin (x + h)}{h} = \lim_{h \to 0} x^{2} [\frac{\sin (x + h) - \sin x}{h}] + \lim_{h \to 0} 2 x \sin (x + h) + \lim_{h \to 0} h \sin (x + h) = \lim_{h \to 0} x^{2} [\frac{2 \sin \frac{(x + h - x)}{2} \cos (\frac{x + h + x}{2})}{h}] + 2 x \sin x + 0 = \lim_{h \to 0} x^{2} [\frac{2 \sin (\frac{h}{2}) \cos (\frac{2 x + h}{2})}{h}] + 2 x \sin x = x^{2} \lim_{(\frac{h}{2}) \to 0} [\frac{\sin (\frac{h}{2})}{(\frac{h}{2})}] \times \lim_{h \to 0} \cos (\frac{2 x + h}{2}) + 2 x \sin x = x^{2} \times 1 \times \cos (\frac{2 x}{x}) + 2 x \sin x = x^{2} \cos x + 2 x \sin x$

Differentiate the following by using first principle - i). tan (2x + 1) ii). √ tan x iii). x2 sin x

Differentiate the following by using first principle -

i). tan (2x + 1)

ii). √ tan x

iii). x²sin x