Rd Sharma XII Vol 1 2021 Solutions for Class 12 Science Maths Chapter 10 Differentiation are provided here with simple step-by-step explanations. These solutions for Differentiation are extremely popular among Class 12 Science 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 2021 Book of Class 12 Science 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 2021 Solutions. All Rd Sharma XII Vol 1 2021 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Page No 10.103:

Question 1:

Find dydx, when

x=at2 and y=2 at

Answer:

We have,  x=at2 and  y=2atdxdt=2at and  dydt=2a dydx=dydtdxdt=2a2at=1t

Page No 10.103:

Question 2:

Find dydx, when

x=a θ+sin θ and y=a 1-cos θ

Answer:

We have, x=aθ+sinθ and y=a1-cosθ

dxdθ=a1+cosθ and dydθ=a sinθ    

dydx=dydθdxdθ=a sinθa1+cosθ=2 sinθ2cosθ22cos2θ=tanθ2

Page No 10.103:

Question 3:

Find dydx, when

x=a cos θ and y=b sin θ

Answer:

We have, x=a cosθ and y=b sinθ dxdθ=-a sinθ and dydθ=b cosθdydx=dydθdxdθ=b cosθ-a sinθ=-bacotθ

Page No 10.103:

Question 4:

Find dydx, when

x=aeθ sin θ-cos θ, y=aeθ sin θ+cos θ

Answer:

We have,  x=aeθsinθ-cosθ and y=aeθsinθ+cosθ

dxdθ=aeθddθsinθ-cosθ+sinθ-cosθddθeθ  and dydθ=aeθddθsinθ+cosθ+sinθ+cosθddθeθdxdθ=aeθcosθ+sinθ+sinθ-cosθeθ and dydθ=aeθcosθ-sinθ+sinθ+cosθeθdxdθ=a2eθ sinθ and dydθ=a2eθ cosθ               

dydθdxdθ=a2eθcosθa2eθsinθ=cotθ

Page No 10.103:

Question 5:

Find dydx, when

x=b sin2 θ and y=a cos2 θ

Answer:

We have, x=b sin2θ and y=a cos2θdxdθ=ddθb sin2θ=2b sinθcosθand, dydθ=ddθa cos2θ=-2a cosθsinθ   dydx=dydθdxdθ=-2a cosθsinθ 2b sinθcosθ=-ab

Page No 10.103:

Question 6:

Find dydx, when

x=a 1-cos θ and y=a θ+sin θ at θ=π2

Answer:

We have, x=a1-cosθ  and  y=aθ+sinθ dxdθ=ddθa1-cosθ=asinθand dydθ=ddθaθ+sinθ=a1+cosθ  dydxθ=π2=dydθdxdθθ=π2=a1+cosθasin θθ=π2=a1+0a=1

Page No 10.103:

Question 7:

Find dydx, when

x=et+e-t2 and y=et-e-t2

Answer:

We have, x=et+e-t2 and y=et-e-t2

dxdt=12ddtet+ddte-t and  dydt=12ddtet-ddte-tdxdt=12et+e-tddt-t and dydt=12et-e-tddte-tdxdt=12et-e-t=y and dydt=12et+e-t=x       dydt=dydtdxdt=xy

Page No 10.103:

Question 8:

Find dydx, when

x=3 at1+t2, and y=3 at21+t2

Answer:

We have, x=3at1+t2
Differentiating with respect to t,
dxdt=1+t2ddt3at-3atddt1+t21+t22                     using quotient ruledxdt=1+t23a-3at2t1+t22dxdt=3a+3at2-6at21+t22dxdt=3a-3at21-t22dxdt=3a1-t21+t22                           ...iand, y=3at21+t2
Differentiating it with respect to t,

dxdt=1+t2ddt3at2-3at2ddt1+t21+t22                     using quotient ruledxdt=1+t26at-3at22t1+t22dxdt=6at+6at3-6at31+t22dxdt=6at1+t22                         ...iiDividing equation ii by i,dydtdxdt=6at1+t22×1+t223a1-t2=2t1-t2

Page No 10.103:

Question 9:

Find dydx, when

x=a cos θ + θ sin θ and y=a sin θ-θ cos θ

Answer:

We have, x=acosθ+θ sinθ and y=asinθ-θ cosθdxdθ =addθcosθ+ddθθ sinθ and dydθ=addθsinθ-ddθθ cosθdxdθ =a-sinθ+θddθsinθ+sinθddθθ and dydθ=acosθ-θddθcosθ+cosθddθθ dxdθ =a-sinθ+θ cosθ and dydθ=acosθ+θ sinθ-cosθdxdθ  =aθ cosθ and dydθ=aθ sinθ dydx=dydθdxdθ=aθ sinθaθ cosθ =tanθ  

Page No 10.103:

Question 10:

Find dydx, when

x=eθ θ+1θ and y=e-θ θ-1θ

Answer:

We have, x=eθθ+1θ
Differentiating it with respect to θ,
dxdθ=eθddθθ+1θ+θ+1θddθeθ                       using product ruledxdθ=eθ1-1θ2+θ2+1θeθdxdθ=eθ1-1θ2+θ2+1θdxdθ=eθθ2-1+θ3+θθ2dxdθ=eθθ3+θ2+θ-1θ2                       ...iand,       y=eθθ-1θ 
Differentiating it with respect to θ using chain rule,

dydθ=e-θddθθ-1θ+θ-1θddθe-θ                       using product ruledydθ=e-θ1+1θ2+θ-1θeθddθ-θdydθ=e-θ1+1θ2+θ-1θe-θ-1dydθ=e-θ1+1θ2-θ+1θdydθ=e-θθ2+1-θ3+θθ2dydθ=e-θ-θ3+θ2+θ+1θ2                     ...iiDividing equation ii by i,dydθdxdθ=e-θθ2-θ3+θ+1θ2×θ2eθθ3+θ2+θ-1        =e-2θθ2-θ3+θ+1θ3+θ2+θ-1 

Page No 10.103:

Question 11:

Find dydx, when

x=2 t1+t2 and y=1-t21+t2

Answer:

We have, x=2t1+t2

dxdt=1+t2ddt2t-2tddt1+t21+t22                          using quotient ruledxdt=1+t22-2t2t1+t22dxdt=2+2t2-4t21+t22dxdt=2-2t21+t22                              ...iand,y=1-t21+t2

dydt=1+t2ddt1-t2-1-t2ddt1+t21+t22dydt=1+t2-2t-1-t22t1+t22dydt=-4t1+t22                          ...iiDividing equation ii by i, we get,dydtdxdt=-4t1+t22×1+t2221-t2dydx=-2t1-t2dydx=-xy                         xy =2t1+t2×1+t21-t2=2t1-t2

Page No 10.103:

Question 12:

Find dydx, when

x=cos-1 11+t2 and y=sin-1 t1+t2, t  R

Answer:

We have, x=cos-111+t2

dxdt=-11-11+t22ddt11+t2dxdt=-11-11+t2-121+t232ddt1+t2dxdt=1+t2121+t2-1×121+t2322tdxdt=tt2×1+t2dxdt=11+t2                                   ...iNow, y=sin-111+t2

dydt=11-11+t22ddt11+t2dydt=11-11+t2-121+t232ddt1+t2dxdt=1+t2121+t2-1×-121+t2322tdxdt=-12t2×1+t22tdxdt=-11+t2                                   ...iiDividing equation ii by i,dydtdxdt=11+t2×1+t2-1dydx=-1

Page No 10.103:

Question 13:

Find dydx, when

x=1-t21+t2 and y=2 t1+t2

Answer:

We have, y=2t1+t2

dydt=1+t2ddt2t-2tddt1+t21+t22                          using quotient ruledydt=1+t22-2t2t1+t22dydt=2+2t2-4t21+t22dydt=2-2t21+t22                              ...iand,x=1-t21+t2

dxdt=1+t2ddt1-t2-1-t2ddt1+t21+t22dxdt=1+t2-2t-1-t22t1+t22dxdt=-4t1+t22                          ...iiDividing equation i by ii, we get,dydtdxdt=21-t21+t22×1+t22-4tdydx=21-t2-4tdydx=t2-12t                      

Page No 10.103:

Question 14:

If x=2 cos θ-cos 2 θ and y=2 sin θ-sin 2 θ, prove that dydx=tan 3 θ2

Answer:

We have, x=2 cosθ- cos2θ

dxdθ=2-sinθ--sin2θddθ2θdxdθ=-2sinθ+2 sin2θdxdθ=2sin2θ-sinθ              ...iand,y=2 sinθ-sin2θ

dydθ=2 cosθ-cos2θddθ2θdydθ=2 cosθ-cos2θ2dydθ=2 cosθ-2 cos2θdydθ=2cosθ-cos2θ              ...iiDividing equation ii by equation i,dydθdxdθ=2cosθ-cos2θ2sin2θ-sinθdydx=cosθ-cos2θsin2θ-sinθdydx=-2sinθ+2θ2sinθ-2θ22cos2θ+θ2sin2θ-θ2                        sinA-sinB=2 cosA+B2sinA-B2and cosA-cosB=-2sinA+B2sinA-B2dydx=-sin3θ2sin-θ2cos3θ2sinθ2dydx=-sin3θ2-sinθ2cos3θ2sinθ2dydx=sin3θ2cos3θ2dydx=tan3θ2

Page No 10.103:

Question 15:

If x=ecos 2 t and y=esin 2 t, prove that dydx=-y log xx log y

Answer:

We have, x=ecos2t and y=esin2t

dxdt=ddtecos2t and dydt=ddtesin2tdxdt=ecos2tddtcos2t and dydt=esin2tddtsin2tdxdt=ecos2t-sin2tddt2t and dydt=esin2tcos2tddt2t            dxdt=-2sin 2tecos2t anddydt=2cos2tesin2t           dydtdxdt=2 cos2tesin2t-2sin2tecos2tdydx=-y logxx logy                                     x=ecos2tlogx=cos2ty=esin2tlogy=sin 2t                    

 

Page No 10.103:

Question 16:

If x=cos t and y=sin t, prove that dydx=13 at t=2 π3

Answer:

We have, x=cost and y=sin t

dxdt=ddtcos t and dydt=ddtsin tdxdt=-sin t and dydt=cos t            dydtdxdt=cos t-sin t=-cot t    Now,  dydxt=2π3=-cot 2π3=13                                                    

Page No 10.103:

Question 17:

If x=at+1t and y=at-1t, prove that dydx=xy

Answer:

We have,  x=at+1t and y=at-1t

dxdt=addtt+1t and dydt=addtt-1tdxdt=a1-1t2 and dydt=a1+1t2  dxdt=at2-1t2 and dydt=at2+1t2                 dydx = dydtdxdt=at2+1t2×t2at2-1dydx =at2+1t×tat2-1dydx =at+1t×1at-1t dydx=xy                                                                               

Page No 10.103:

Question 18:

If x=sin-1 2 t1+t2 and y=tan-1 2 t1-t2,-1<t<1, prove that dydx=1

Answer:

We have, x=sin-12t1+t2Put t=tanθ-1<tanθ<1-π4<θ<π4-π2<2θ<π2x=sin-12 tanθ1+tan2θ x=sin-1sin2θx=2θ                                -π2<2θ<π2x=2tan-1t                   t=sinθ

dxdt=21+t2           ...iNow, y=tan-12t1-t2put t=tanθ  y=tan-12 tanθ1-tan2θ  y=tan-1tan 2θ               y=2θ                                     -π2<2θ<π2 y=2 tan-1t                              t=tanθ    

dydt=21+t2        ...iiDividing equation ii by i,dydtdxdt=21+t2×1+t22dydx=1

Page No 10.103:

Question 19:

If x=sin3 tcos 2 t, y=cos3 tcost 2 t, find dydx

Answer:

We have, x=sin3tcos2t  and  y=cos3tcos2tdxdt=ddtsin3tcos2t dxdt=cos2tddtsin3t-sin3tddtcos2tcos2t              Using quotient ruledxdt=cos2t3sin2tddtsint-sin3t×12cos2tddtcos 2tcos2t       dxdt=3cos2tsin2t cost-sin3t2cos2t-2 sin2tcos 2t      dxdt=3cos2t sin2t cost+sin3t sin2tcos2tcos2tNow, dydt=ddtcos3tcos2t       dydt=cos2tddtcos3t-cos3tddtcos2tcos2t              Using quotient rule      dydt=cos2t3cos2tddtcost-cos3t×12cos2tddtcos 2tcos2t       dydt=3cos2tcos2t -sint-cos3t2cos2t-2 sin2tcos 2t      dydt=-3cos2t cos2t sint+cos3t sin2tcos2tcos2tdydx=dydtdxdt=-3cos2t cos2t sint+cos3t sin2tcos2tcos2t×cos2tcos2t3cos2t sin2t cost+sin3t sin2t    dydx=-3cos2t cos2t sint+cos3t sin2t3cos2t sin2t cost+sin3t sin2t    dydx=sint cost-3cos2t cost+2cos3tsint cost3cos2t sint+2sin3t    dydx=-32cos2t-1cost+2cos3t31-2sin2tsint+2sin3t                    cos2t=2cos2t-1cos2t=1-2sin2t    dydx=-4cos3t+3cost3sint-4sin3t    dydx=-cos3tsin3t                                   cos3t=4cos3t-3costsin3t=3sint-4sin3tdydx=-cot3t

Page No 10.103:

Question 20:

If x=t+1ta, y=at+1t, find dydx

Answer:

We have, x=t+1ta

dxdt=ddtt+1ta                   dxdt=at+1ta-1ddtt+1tdxdt=at+1ta-11-1t2            ...i             and, y=at+1t

dydt=ddtat+1t         dydt=at+1t× logaddtt+ttdydt=at+1t× loga1-1t2          ...iiDividing equation ii by i,dydtdxdt=at+1t× loga1-1t2  at+1ta-11-1t2dydx=at+1t× logaat+1ta-1

Page No 10.103:

Question 21:

If x=a 1+t21-t2 and y=2t1-t2, find dydx

Answer:

We have, x=a1+t21-t2

dxdt=a1-t2ddt1+t2-1+t2ddt1-t21-t22                    Using quotient ruledxdt=a1-t22t-1+t2-2t1-t22dxdt=a2t-2t3+2t+2t31-t22dxdt=4at1-t22                ...iand, y=2t1-t2

dydt=21-t2ddtt-tddt1-t21-t22                    Using quotient ruledydt=21-t21-t-2t1-t22dydt=21-t2+2t21-t22dydt=21+t21-t22                ...iiDividing equation ii by i,dydtdxdt=21+t21-t22  ×1-t224atdydx=1+t22at



Page No 10.104:

Question 22:

If x=10 t-sin t, y=12 1-cos t, find dydx.

Answer:

We have, x=10t-sint and y=121-costdxdt=ddt10t-sint anddydt=ddt121-costdxdt=10ddtt-sint and dydt=12ddt1-costdxdt=101-cost and dydt=120--sint=12 sintdydx=dydtdxdt=12 sint101-costdydx=12×2sint2cost210×2sin2t2dydx=65cott2

Page No 10.104:

Question 23:

If x=a θ-sin θ and, y=a 1+cos θ, find dydx at θ=π3.

Answer:

We have, x=aθ-sinθ  and  y=a1+cosθ dxdθ=ddθaθ-sinθ and dydθ=ddθa1+cosθdxdθ=a1-cosθ and dydθ=a-sinθ dydx=dydθdxdθ=-a sinθa1-cosθNow, dydxθ=π3=-sinπ31-cosπ3=-321-12=-3

Page No 10.104:

Question 24:

If x=asin2t1+cos2t and y=bcos2t1-cos2t, show that at t=π4, dydx=bat=π4, dydx=ba

Answer:

x=asin2t1+cos2t and y=bcos2t1-cos2tdxdt=2acos2t1+cos2t+2asin2t1-cos2t and dydt=-2bsin2t1-cos2t+2bcos2t1+cos2t dxdt=2acos2t+cos22t+sin2t-sin2tcos2t and dydt=2b-sin2t+sin2tcos2t+cos2t+cos22tdydt=dydtdxdt=-2b-sin2t+sin2tcos2t+cos2t+cos22t2acos2t+cos22t+sin2t-sin2tcos2tdydtt=π4=-2b-sinπ2+sinπ2cosπ2+cosπ2+cos2π22acosπ2+cos2π2+sinπ2-sinπ2cosπ2=ba

Page No 10.104:

Question 25:

If x=cost3-2cos2t, y=sint3-2sin2t find the value of dydx at t=π4

Answer:

x=cost3-2cos2t and y=sint3-2sin2tdxdt=-sint3-2cos2t+cost4costsint and dydt=cost3-2sin2t+sint-4sintcostdxdt=-3sint+6sintcos2t and dydt=3cost-6sin2tcostdxdt=-3sint1-2cos2t and dydt=3cost1-2sin2tdxdt=3sintcos2t and dydt=3costcos2tdydx=dydtdxdt=3costcos2t3sintcos2t=cottNow, dydxt=π4=cotπ4=1

Page No 10.104:

Question 26:

If x=1+logtt2, y=3+2logtt, find dydx

Answer:

x=1+logtt2 and y=3+2logttdxdt=t-2t-2tlogtt4 and dydt=2-3-2logtt2dxdt=-1-2logtt3 and dydt=-1-2logtt2dydx=dydtdxdt=-1-2logtt2-1-2logtt3=t

Page No 10.104:

Question 27:

sinx=2t1+t2, tany=2t1-t2, find dydx

Answer:

sinx=2t1+t2 and tany=2t1-t2x=sin-12t1+t2 and y=tan-12t1-t2x=2tan-1t and y=2tan-1tdxdt=2t1+t2 and dydt=2t1+t2dydx=dydtdxdt=2t1+t22t1+t2=1

Page No 10.104:

Question 28:

Write the derivative of sinx with respect to cosx

Answer:

Let u=sinx and v=cosxdudx=cosx and dvdx=-sinxdudxdvdx=cosx-sinxdudv=-cotx

Page No 10.104:

Question 29:

If x = a (2θ –  sin 2θ) and y = a (1 – cos 2θ), find dydx when θ=π3.

Answer:

Given values are:
x=a2θ-sin2θandy=a1-cos2θ

Applying parametric differentiation

dxdθ= 2a − 2acos2θ

dydθ= 0 + 2asin2θ

dydxdydθ×dθdx=sin2θ1- cos2θ

Now putting the value of θπ3
dydxθ=π3=sin2π31-cos2π3                =321+12                =3232=13
So, dydx is 13 at θ=π3.



Page No 10.112:

Question 1:

Differentiate x2 with respect to x3

Answer:

Let u=x2  and v=x3dudx=2x and dvdx=3x2     dudv=dudxdvdx=2x3x2=23x

Page No 10.112:

Question 2:

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

Answer:

Let u=log1+x2 and v=tan-1x dudx=11+x2ddx1+x2 =2x1+x2 and dvdx=11+x2                   dudv=dudxdvdx=2x1+x2×1+x21=2x
 

Page No 10.112:

Question 3:

Differentiate (log x)x with respect to log x

Answer:

Let u=logxx
Taking log on both sides,
log u=loglogxxlog u=x loglogx           

1ududx=xddxloglogx+loglogxddxx1ududx=x1logxddxlogx+loglogx1dudx=uxlogx1x+log logxdudx=logxx1logx+log logx                   ..iAgain, let v=logxdvdx=1x                     ...iiDividing equation i by ii,we getdudxdvdx=logxx1logx+log logx1xdudv=logxx1+logxlog logxlogx1xdudv=xlogxx-11+logx × log logx

Page No 10.112:

Question 4:

Differentiate sin-1  1-x2 with respect to cos-1 x, if
(i) x  0, 1
(ii) x  -1, 0

Answer:

i Let,    u=sin-11-x2Put    x=cosθ     u=sin-11-cos2θ    u=sin-1sinθ                ...iAnd, v=cos-1x               ...iiNow,   x0,1    cosθ0,1      θ0,π2So, from equation i,   u=θ                   Since, sin-1sinθ=θ if θ-π2,π2u=cos-1x            Since, cosθ=x 

Differentiating it with respect to x,

dudx=-11-x2              ...iiifrom equation ii,v=cos-1x

Differentiating it with respect to x,

dvdx=-11-x2           ...ivDividing equation iii by iv,dudxdvdx=-11-x2×1-x2-1dudx=1


ii  Let,    u=sin-11-x2Put    x=cosθ     u=sin-11-cos2θ     u=sin-1sinθ                ...iAnd, v=cos-1x                      ...iiNow,   x-1,0    cosθ-1,0      θπ2,πSo, from equation i,  u=π-θ                   Since, sin-1sinθ=π-θ if θπ2,3π2 u=π-cos-1x            Since, x=cosθ 

Differentiating it with respect to x,

dudx=0--11-x2   dudx=  11-x2           ...iiifrom equation ii,v=cos-1x

Differentiating it with respect to x,

dvdx=-11-x2           ...ivDividing equation iii by iv,dudxdvdx=11-x2×1-x2-1dudx=-1

Page No 10.112:

Question 5:

Differentiate sin-1 4x 1-4x2 with respect to 1-4x2, if

(i) x -12 2, 12 2

(ii) x  12 2, 12

(iii) x  -12, -12 2

Answer:

i    Let,   u=sin-14x1-4x2       put    2x=cosθ             u=sin-12×cosθ1-cos2θ             u=sin-12cosθ sinθ              u=sin-1sin 2θ                      ...i     Let,     v=1-4x2                            ...iiHere,         x-122, 122    2x-12, 12  θπ4,3π4So, from equation i,    u=π-2θ                            Since, sin-1sinθ=π-θ ,if θπ2, π u=π-2 cos-12x            Since, 2x=cosθ                   

Differentiating it with respect to x,

dudx=0-2-11-2x2ddx2xdudx=21-4x22dudx=41-4x2                        ...iiifrom equation iidvdx=-4x1-4x2but, x-12,-122dvdx=-4-x1-4-x2dvdx=4x1-4x2                         ...ivDiferentiating equation ii with respect to x,dvdx=121-4x2ddx1-4x2dvdx=121-4x2-8xdvdx=-4x1-4x2                     ...vDividing equation iii by vdudxdvdx=41-4x2 ×1-4x2-4xdudv=-1x


ii    Let,   u=sin-14x1-4x2       put  2x=cosθ                 u=sin-12×cosθ1-cos2θ         u=sin-12cosθ sinθ           u=sin-1sin 2θ                      ...i     Let,     v=1-4x2                         ...iiHere,         x122,12    2x12,1  cosθ12,1  θ0,π4So, from equation i,    u=2θ                            Since, sin-1sinθ=θ ,if θ-π2,π2 u=2 cos-12x            Since, 2x=cosθ                   

Differentiate it with respect to x,

dudx=2-11-2x2ddx2xdudx=-21-4x22dudx=-41-4x2                          ...iiiDiferentiating equation ii with respect to x,dvdx=121-4x2ddx1-4x2dvdx=121-4x2-8xdvdx=-4x1-4x2                            ...ivDividing equation iii by ivdudxdvdx=-41-4x2 ×1-4x2-4xdudv=1x


iii    Let,   u=sin-14x1-4x2       put,  2x=cosθ          u=sin-12×cosθ1-cos2θ          u=sin-12cosθ sinθ           u=sin-1sin 2θ                ...i     Let,     v=1-4x2                   ...iiHere,         x-12,-122    2x-1,-12   θ3π4,πSo, from equation i,    u=π-2θ                            Since, sin-1sinθ=π-θ ,if θπ2,3π2 u=π-2 cos-12x            Since, 2x=cosθ                   


Differentiate it with respect to x,

dudx=0-2-11-2x2ddx2xdudx=21-4x22dudx=41-4x2                         ...iiifrom equation ii,dvdx=-4x1-4x2but, x-12,-122dvdx=-4-x1-4-x2dvdx=4x1-4x2                     ...ivDividing equation iii by ivdudxdvdx=41-4x2 ×1-4x24xdudv=1x

Page No 10.112:

Question 6:

Differentiate tan-1 1+x2-1x with respect to sin-1 2x1+x2, if -1<x<1, x0.

Answer:

Let,     u=tan-11+x2-1xput     x=tanθ       u=tan-11+tan2θ-1tanθ      u=tan-1secθ-1tanθ     u=tan-11-cosθsinθ       u=tan-12sin2θ22sinθ2cosθ2    u=tan-1 tanθ2                     ...iAnd,       v=sin-12x1+x2    v=sin-12tanθ1+tan2θ     v=sin-1sin2θ                      ...iiHere,     -1<x<1 -1<tanθ<1  -π4<θ<π4                       ...A So, from equation i,u=θ2                        Since, tan-1tanθ=θ, if θ-π2,π2 u=12tan-1x             since, x=tanθ

Differentiating it with respect to x,

dudx=1211+x2dudx=121+x2         ...iNow, from equation ii and A,v=2θ                            Since, sin-1sinθ=θ, if θ-π2,π2v=2tan-1x                  Since, x=tanθ

Differentiating it with respect to x,

dvdx=211+x2          ...ivdividing equation iii by iv,dudxdvdx=121+x2×1+x22dudv=14

Page No 10.112:

Question 7:

Differentiate sin-1 2x 1-x2 with respect to sec-1 11-x2, if
(i) x 0, 12

(ii) x 12, 1

Answer:

i Let,    u=sin-12x1-x2     Put    x=sinθ          u=sin-12sinθ1-sin2θ          u=sin-12 sinθ cosθ            u=sin-1sin2θ                     ...iAnd,   Let        v=sec-111-x2            v=sec-111-sin2θ               v=sec-11cosθ              v=sec-1secθ               v=cos-111cosθ                       Since, sec-1x=cos-11x             v=cos-1cosθ                  ...iiHere,           x0,12   sinθ0,12          θ0,π4So, from equation i,    u=2θ                                 Since, sin-1sinθ=θ, if θ-π2,π2 Let, u=2sin-1x                         Since, x=sinθ

Differentiating it with respect to x,

dudx=211-x2dudx=21-x2             ...iiiAnd, from equation ii,v=θ                                 Since, cos-1cosθ=θ, if θ0,πv=sin-1x                         Since, x=sinθ

Differentiating it with respect to x,

dvdx=11-x2            ...ivdividing equation iii by iv,dudxdvdx=21-x2×1-x21dudv=2


ii Let,    u=sin-12x1-x2     Put    x=sinθ          u=sin-12sinθ1-sin2θ          u=sin-12 sinθ cosθ            u=sin-1sin2θ                    ...iAnd,   Let,      v=sec-111-x2            v=sec-111-sin2θ               v=sec-11cosθ              v=sec-1secθ               v=cos-111cosθ                       Since, sec-1x=cos-11x             v=cos-1cosθ                ...iiHere,           x12,1   sinθ12,1          θπ4,π2So, from equation i,    u=2θ                                 Since, sin-1sinθ=θ, if θ-π2,π2 Let, u=2sin-1x                         Since, x=sinθ

Differentiating it with respect to x,

dudx=211-x2dudx=21-x2            ...iiiAnd, from equation ii,v=θ                                 Since, cos-1cosθ=θ, if θ0,πv=sin-1x                         Since, x=sinθ

Differentiating it with respect to x,

dvdx=11-x2            ...ivdividing equation iii by iv,dudxdvdx=21-x2×1-x21dudv=2

Page No 10.112:

Question 8:

Differentiate cos xsin x with respect to sin xcos x.

Answer:

Let,     u=cosxsinx

Taking log on both sides,

log u=logcosxsinxlog u=sinx logcosx

Differentiating it with respect to x using chain rule,

1ududx=sinxddxlog cosx+log cosxddxsinx             using product rule1ududx=sinx1cosxddxcosx+log cosxcosxdudx=utanx×-sinx+log logxcosxdudx=cosxsinxcosx logcosx-sinx tanx                             ...iLet,   v=sinxcosx

Taking log on both sides,

log v=logsinxcosxlog v=cosx logsinx

Differentiating it with respect to x using chain rule,

1vdvdx=cosxddxlogsinx+logsinxddxcosx            using product rule1vdvdx=cosx1sinxddxsinx+logsinx-sinxdvdx=vcotxcosx-sinx logsinxdvdx=sinxcosxcotxcosx-sinx logsinxdividing equationi by ii,dudv=cosxsinxcosx logcosx-sinx tanxsinxcosxcotxcosx-sinx logsinx

Page No 10.112:

Question 9:

Differentiate sin-1 2x1+x2 with respect to cos-1 1-x21+x2, if 0<x<1

Answer:

Let,   u=sin-12x1+x2Put   x=tanθ    u=sin-12tanθ1+tan2θ    u=sin-1sin2θ               ...iLet   v=cos-11-x21+x2    v=cos-11-tan2θ1+tan2θ    v=cos-1cos2θ          ...iiHere, 0<x<1     0<tanθ<1     0<θ<π4So, from equation i,u=2θ                            Since, sin-1sinθ=θ , if θ-π2,π2u=2tan-1x                  Since , x=tanθ

Differentiating it with respect to x,

dudx=21+x2                ...iiifrom equation ii,v=2θ                         Since, cos-1cosθ=θ, if θ0,πv=2tan-1x               Since, x=tanθ

Differentiating it with respect to x,

dvdx=21+x2                ...ivDividing equation iii by iv,dudxdvdx=21+x2×1+x22dudv=1



Page No 10.113:

Question 10:

Differentiate tan-1 1+ax1-ax with respect to 1+a2 x2

Answer:

Let,   u=tan-11+ax1-axPut   ax=tanθ    u=tan-11+tanθ1-tanθ    u=tan-1tanπ4+tanθ1-tanπ4tanθ    u=tan-1tanπ4+θ    u=π4+θ    u=π4+tan-1ax                  Since, tanθ=ax               

Differentiating it with respect to x,

dudx=0+11+ax2ddxax  dudx=a1+a2x2                  ...i              Now,Let,   v=1+a2x2

Differentiating it with respect to x,

dvdx=121+a2x2ddx1+a2x2dvdx=121+a2x22a2xdvdx=a2x1+a2x2                ...ii            Dividing equation i by ii,dudxdvdx=a1+a2x2×1+a2x2a2xdudv=1ax1+a2x2

Page No 10.113:

Question 11:

Differentiate sin-1 2x 1-x2 with respect to tan-1 x1-x2, if -12<x<12

Answer:

Let,   u=sin-12x1-x2Put   x=sinθ    u=sin-12sinθ1-sin2θ    u=sin-12 sinθcosθ     u=sin-1sin2θ                 ...iLet  v=tan-1x1-x2   v=tan-1sinθ1-sin2θ     v=tan-1sinθcosθ   v=tan-1tanθ                  ...iiHere,    -12<x<12       -12<sinθ<12       -π4<θ<π4So, from equation i,u=2θ                      Since, sin-1sinθ=θ, if θ-π2,π2u=2sin-1x             Since, x=sinθ

Differentiating it with respect to x,

dudx=21-x2            ...iiifrom equation ii,v=θ                           Since, tan-1tanθ=θ, if θ-π2,π2v=sin-1x                  Since, x=sinθ

Differentiating it with respect to x,

dvdx=11-x2            ...ivDividing equation iii by iv,dudxdvdx=21-x2×1-x21dudv=2

Page No 10.113:

Question 12:

Differentiate tan-1 2x1-x2 with respect to cos-1 1-x21+x2, if 0<x<1

Answer:

Let,     u=tan-12x1-x2Put     x=tanθ      u=tan-12tanθ1-tan2θ      u=tan-1tan2θ               ...ilet,      v=cos-11-x21+x2     v=cos-11-tan2θ1+tan2θ      v=cos-1cos2θ                ... iiHere,  0<x<1       0<tanθ<1       0<θ<π4So, from equation i,u=2θ                          Since, tan-1tanθ=θ, if θ-π2,π2u=2tan-1x                 Since, x=tanθ


differentiating it with respect to x,

dudx=21+x2                 ...iiiFrom equation ii,v=θ                        Since, cos-1cosθ=θ, if θ0,πv=2tan-1x            Since, x=tanθ


Differentiating it with respect to x,


dvdx=21+x2                ...ivDividing equation iii by iv,dudxdvdx=21+x2×1+x22dudv=1

Page No 10.113:

Question 13:

Differentiate tan-1 x-1x+1 with respect to sin-1 3x-4x3, if -12<x<12

Answer:

Let,     u=tan-1x-1x+1Put     x=tanθ      u=tan-1tanθ-1tanθ+1      u=tan-1tanθ-tanπ41+tanθ tanπ4          u=tan-1tanθ-π4               ...i    Here,  -12<x<12       -12<tanθ<12       -tan-112<θ<tan-112So, from equation i,u=θ-π4                          Since, tan-1tanθ=θ, if θ-π2,π2u=tan-1x -π4                Since, x=tanθ

differentiating it with respect to x,

dudx=11+x2-0  dudx=11+x2               ...ii              And,Let,     v=sin-13x-4x3Put     x=sinθ      v=sin-13sinθ-4sin3θ      v=sin-1sin3θ          ...iiiNow,  -12<x<12    -12<sinθ<12    -16<θ<π6So, from equation iii,v=3θ                           Since, sin-1sinθ=θ, if θ-π2,π2v=3sin-1x                  Since, x=sinθ

Differentiating it with respect to x,

dvdx=31-x2                ...ivDividing equation iii by iv,dudxdvdx=11+x2×1-x23dudv=1-x231+x2

Page No 10.113:

Question 14:

Differentiate tan-1 cos x1+sin x with  respect to sec-1 x

Answer:

Let,    u=tan-1cosx1+sinx    u=tan-1tanπ4-x2    u=π4-x2

Differentiating it with respect to x,

dudx=0-12dudx=-12                ...iLet,    v=sec-1x

Differentiating it with respect to x,

dvdx=1xx2-1           ...iiDividing equation i by ii,dudxdvdx=-12×xx2-11dudv=-xx2-12

Page No 10.113:

Question 15:

Differentiate sin-1 2x1+x2 with respect to tan-1 2 x1-x2, if -1<x<1

Answer:

Let,    u=sin-12x1+x2Put     x=tanθ θ=tan-1x,      u=sin-12tanθ1+tan2θ      u=sin-1sin2θ                 ...iLet,    v=tan-12x1-x2      v=tan-12tanθ1-tan2θ      v=tan-1tan2θ                ...iiHere,  -1<x<1       -1<tanθ<1       -π4<tanθ<π4So, from equation i,u=2θ                          Since, sin-1sinθ=θ, if θ-π2,π2u=2tan-1x                


Differentiating it with respect to x,

dudx=21+x2             ...iiifrom equation ii,v=2θ                    Since, tan-1tanθ=θ , if θ-π2,π2v=2tan-1x

Differentiating it with respect to x,

dvdx=21+x2                 ...ivDividing equation iii by iv,dudxdvdx=21+x2       ×1+x22dudv=1

Page No 10.113:

Question 16:

Differentiate cos-1 4x3-3x with respect to tan-1 1-x2x, if 12<x<1

Answer:

Let,    u=cos-14x3-3xPut,    x=cosθ      θ=cos-1xNow,   u=cos-14cos3θ-3cosθ       u=cos-1cos3θ                     ...i Let,     v=tan-11-x2x      v=tan-11-cos2θcosθ        v=tan-1sinθcosθ      v=tan-1tanθ                          ...iiHere,        12<x<1   12<cosθ<1   0<θ<π3So, from equation i,u=3θ                              Since, cos-1cosθ=θ, if θ0,πu=3cos-1x

Differentiating it with respect to x,

dudx=-31-x2             ...iiiFrom equation ii,v=θ                     Since, tan-1tanθ=θ, if θ-π2,π2v=cos-1x

Differentiating it with respect to x,

dvdx=-11-x2               ...ivDividing equation iii by iv,dudxdvdx=-31-x2-1-x21dudv=3

Page No 10.113:

Question 17:

Differentiate tan-1 x1-x2 with respect to sin-1 2x 1-x2, if -12<x<12

Answer:

 Let,     u=tan-1x1-x2Put    x=sinθ      θ=sin-1x      u=tan-1sinθ1-sin2θ        u=tan-1sinθcosθ      u=tan-1tanθ                 ...iAndLet,    v=sin-12x1-x2           v=sin-12sinθ1-sin2θ           v=sin-12 sinθcosθ           v=sin-1sin2θ                 ...iiHere,        -12<x<12   -12<sinθ<12   -π4<θ<π4So, from equation i,u=θ                              Since, tan-1tanθ=θ, if θ-π2,π2u=sin-1x

Differentiating it with respect to x,

dudx=11-x2             ...iiifrom equation ii,v=2θ                     Since, sin-1sinθ=θ, if θ-π2,π2 v=2sin-1x 

Differentiating it with respect to x,

dvdx=21-x2               ...ivDividing equation iii by iv,dudxdvdx=11-x21-x22dudv=12

Page No 10.113:

Question 18:

Differentiate sin-1 1-x2 with respect to cot-1 x1-x2, if 0<x<1

Answer:

Let,    u=sin-11-x2Put    x=cosθ      θ=cos-1xWe get,     u=sin-1sinθ                     ...i Let,     v=cot-1x1-x2      v=cot-1cosθ1-cos2θ        v=cot-1cosθsinθ      v=cot-1cotθ                       ...iiHere,        0<x<1   0<cosθ<1   0<θ<π2So, from equation i,u=θ                              Since, sin-1sinθ=θ, ifθ-π2,π2u=cos-1x

Differentiating it with respect to x,

dudx=-11-x2             ...iiiFrom equation ii,v=θ                     Since, cot-1cotθ=θ, if θ0,πv=cos-1x

Differentiating it with respect to x,

dvdx=-11-x2               ...ivDividing equation iii by iv,dudxdvdx=-11-x21-x2-1dudv=1

Page No 10.113:

Question 19:

Differentiate sin-1 2 ax 1-a2 x2 with respect to 1-a2 x2, if-12<ax<12

Answer:

 Let,     u=sin-12ax1-a2x2Put    ax=sinθ       θ=sin-1ax     u=sin-12sinθ1-sin2θ       u=sin-12 sinθcosθ     u=sin-1sin2θ                         ...iAndLet,           v=1-a2x2Differentiating it with respect to x,dvdx=121-a2x2 ×ddx1-a2x2       dvdx=0-2a2x21-a2x2    dvdx=-a2x1-a2x2                            ...iiHere,        -12<ax<12   -12<sinθ<12   -π4<θ<π4So, from equation i,u=2θ                              Since, sin-1sinθ=θ, if θ-π2,π2u=2sin-1x

Differentiating it with respect to x,

dudx=2×11-ax2ddxax  dudx=21-a2x2a  dudx=2a1-a2x2       ...iii      

Dividing equation iii by ii,dudxdvdx=2a1-a2x21-a2x2-a2xdudv=-2ax

Page No 10.113:

Question 20:

Differentiate tan-1 1-x1+x with respect to 1-x2, if-1<x<1

Answer:

Let,   u=tan-11-x1+xPut    x=tanθ     θ=tan-1x    u=tan-11-tanθ1+tanθ    u=tan-1tanπ4-θ               ...iHere,             -1<x<1     -1<tanθ<1     -π4<θ<π4     π4>-θ>π4     -π4<-θ<π4     0<π4-θ<π2So, from equation i,u=π4-θ                               Since, tan-1tanθ=θ, if θ-π2,π2u=π4-tan-1x

Differentiating it with respect to x,

dudx=0-11+x2dudx=-11+x2           ...iiAnd  let,      v=1-x2

Differentiating it with respect to x,

dvdx=121-x2×ddx1-x2dvdx=121-x2-2xdvdx=-x1-x2                     ...iiiDividing equation ii by iii,dudxdvdx=-11+x2×1-x2-xdudv=1-x2x1+x2

Page No 10.113:

Question 21:

Differentiate sinx with respect to ecos x.

Answer:

Let μ=sin2x and v=ecosx
Now,
μ=sin2xdμdx=2sinx cosx
Also,
v=ecosxdvdx=ecosx-sinx
dμdv=dvdxdvdx=2sinx cosxecosx-sinx=-2cosxecosx=-2cosxe-cosx



Page No 10.117:

Question 1:

If f (x) = logx2 (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, fx=logx2logxfx=loglogxlogx2       fx=loglogx2 logxf'x=12×ddxloglogx logxf'x=12×1logx×1x×logx-loglogxx logx2f'x=12×1x-loglogxx logx2f'e=12×1e-loglogee loge2       Putting x=ef'e=12×1e1f'e=12e

Page No 10.117:

Question 2:

The differential coefficient of f (log x) w.r.t. x, where f (x) = log x is
(a) xlog x

(b) log xx

(c) x log x-1

(d) none of these

Answer:

(c) x log x-1

We have,
fx=log xflogx=loglogxf'logx=1logxddxlogxf'logx=1x logxf'logx=x logx-1

Page No 10.117:

Question 3:

The derivative of the function cot-1 cos 2 x1/2 at x=π/6 is
(a) (2/3)1/2
(b) (1/3)1/2
(c) 31/2
(d) 61/2

Answer:

(a) (2/3)1/2
We have,  y=cot-1cos 2x

dydx=-11+cos 2xddxcos 2xdydx=-12 cos2x×12cos 2xddxcos 2xdydx=-12 cos2x×12cos 2x×-2sin 2xdydx=sin2xcos2x × 2cos2xdydx=2 sinx cosxcos2x × 2cos2xdydx=tanxcos2xSo, at x=π6, we getdydxx=π6=tanπ6cos2π6=1312=2312

Page No 10.117:

Question 4:

Differential coefficient of sec sec tan-1 x is
(a) x1+x2

(b) x 1+x2

(c) 11+x2

(d) x1+x2

Answer:

(d) x1+x2

We have, y=sectan-1xdydx=sectan-1x tantan-1x×ddxtan-1xdydx=sectan-1x tantan-1x×11+x2dydx=yx1+x2dydx=x1+x2 y
This is the equation of differential equation which have coefficient x1+x2.

Page No 10.117:

Question 5:

If fx=tan-1 1+sin x1-sin x, 0xπ/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-11+sinx1-sinx y=tan-11-cosπ2+x1+cosπ2+x y=tan-12 sin2π4+x22 cos2π4+x2  y=tan-1tanπ4+x2=π4+x2dydx=12

Page No 10.117:

Question 6:

If y=1+1xx, then dydx=

(a) 1+1xx 1+1x-1x+1

(b) 1+1xx log 1+1x

(c) x+1xx log x+1-xx+1

(d) x+1xx log 1+1x+1x+1

Answer:

(a) 1+1xx 1+1x-1x+1

Let y=1+1xxTaking log on both sides,log y=x log1+1x1ydydx=xddxlog1+1x+log1+1xddxx            1ydydx=x11+1xddx1+1x+log1+1x1ydydx=x × xx+1-1x2+log1+1x1ydydx=x2x+1×-1x2+log1+1x1ydydx=-1x+1+log1+1xdydx=y-1x+1+log1+1xdydx=1+1xxlog1+1x-1x+1



Page No 10.118:

Question 7:

If xy=ex-y, then dydx is
(a) 1+x1+log x

(b) 1-log x1+log x

(c) not defined

(d) log x1+log x2

Answer:

(d) log x1+log x2

We have, xy=ex-yTaking log on both sides we get, y logx=x-yloge ey logx=x-yy logx+y=xy1+logx=xy=x1+logx
dydx=1+logx × 1-x ×0+1x1+logx2dydx=1+logx -11+logx2dydx=logx1+logx2

Page No 10.118:

Question 8:

Given fx=4x8, then

(a) f'12=f'-12

(b) f12=-f'-12

(c) f-12=f-12

(d) f12=f'-12

Answer:

 c  f-12=f-12We have, fx=4x8f'x=32x7Now,  f12=4128=41256=164 f-12=4-128=41256=164f'12=32127=321128=14f'-12=32-127=-321128=-14

Page No 10.118:

Question 9:

If x=a cos3 θ, y=a sin3 θ, then 1+dydx2=
(a) tan2 θ
(b) sec2 θ
(c) sec θ
(d) sec θ

Answer:

(d) sec θ

We have, x=a cos3θdxdθ=addθcos3θdxdθ=3acos2θddθcosθdxdθ=-3acos2θsinθ                  ...1and,  y=a sin3θdydθ=addθsin3θdydθ=3a sin2θddθsinθdydθ=3a sin2θ cosθ                     ...2Dividing 2 by 1, we get, dydθdxdθ=3a sin2θ cosθ   -3acos2θsinθdydx=sinθ-cosθdydx=-tanθNow, 1+dydx2=1+tan2θ=sec2θ=secθ

Page No 10.118:

Question 10:

If y=sin-1 1-x21+x2, then dydx=

(a) -21+x2

(b) 21+x2

(c) 12-x2

(d) 22-x2

Answer:

(a) -21+x2

Let y=sin-11-x21+x2Differentiating with respect to x using chain rule, we get,dydx=11-1-x21+x22ddx1-x21+x2dydx=1+x21+x22-1-x221+x2ddx1-x2-1-x2ddx1+x21+x22               using quotient ruledydx=1+x21+x22-1-x221+x2-2x-1-x22x1+x22dydx=1+x22x-2x-2x3-2x+2x31+x22dydx=-4x2x1+x2dydx=-21+x2

Page No 10.118:

Question 11:

The derivative of sec-1 12 x2+1 w.r.t. 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-1α is not defined for α-1, 1Here for x=-13, 12x2+1=911-1, 1 sec-112x2+1 is not defined at  x=-13 Derivative of sec-112x2+1 does not exist at x=-13

Page No 10.118:

Question 12:

For the curve x+y=1, dydx at 1/4, 1/4 is
(a) 1/2
(b) 1
(c) −1
(d) 2

Answer:

(c) −1

We have, x+y=1Differentiating with respect to x, we get,12x+12ydydx=012ydydx=-12xdydx=-12x×2y1dydx=-yxNow, dydx14,14=-1414=-1

Page No 10.118:

Question 13:

If sin x+y=log x+y, then dydx=
(a) 2
(b) − 2
(c) 1
(d) − 1]

Answer:

(d) − 1

We have, sinx+y=logx+ycosx+y1+dydx=1x+y1+dydxcosx+y+cosx+ydydx=1x+y+1x+ydydxcosx+ydydx-1x+ydydx=1x+y-cosx+ycosx+y-1x+ydydx=1x+y-cosx+y-1x+y-cosx+ydydx=1x+y-cosx+ydydx=-1

Page No 10.118:

Question 14:

Let =sin-1 2 x1+x2 and V=tan-1 2 x1-x2, then ddV=
(a) 1/2
(b) x
(c) 1-x2x2-4
(d) 1

Answer:

(d) 1
We have, u=sin-12x1+x2 and v=tan-12x1-x2dudx=21+x2  and dvdx=21+x2   

dudv=dudxdvdx=21+x2×1+x22=1

Page No 10.118:

Question 15:

ddx tan-1 cos x1+sin x equals

(a) 1/2
(b) − 1/2
(c) 1
(d) − 1

Answer:

(b) − 1/2

Let  u=tan-1cosx1+sinx   u=tan-1cos2x2-sin2x2cos2x2+sin2x2+2sinx2cosx2   u=tan-1cosx2-sinx2cosx2+sinx2cosx2+sinx22   u=tan-1cosx2-sinx2cosx2+sinx2  u=tan-11-tanx21+tanx2  u=tan-1tanπ4-tanx21+tanπ4×tanx2  u=tan-1tanπ4-x2  u=π4-x2

dudx=0-12dudx= -12



Page No 10.119:

Question 16:

ddx log ex x-2x+23/4 equals

(a) x2-1x2-4

(b) 1

(c) x2+1x2-4

(d) exx2-1x2-4

Answer:

(a) x2-1x2-4

Let y=ddxlogexx-2x+234 y=ddxxloge+34logx-2x+2 y=ddxx+34logx-2x+2dydx=1+34x-2x+2×x+2×1-x-2×1x+22dydx=1+3x+24x-2×x+2-x+2x+22dydx=1+3x+24x-2×4x+2dydx=1+3x2-4dydx=x2-4+3x2-4dydx=x2-1x2-4

Page No 10.119:

Question 17:

If y=sin x+y, then dydx=

(a) sin x2 y-1

(b) sin x1-2 y

(c) cos x1-2 y

(d) cos x2 y-1

Answer:

(d) cos x2 y-1

We have, y=sinx+ySquaring both sides, we get,y2=sinx+yy2-y=sinx2ydydx-dydx=cosxdydx2y-1=cosxdydx=cosx2y-1

Page No 10.119:

Question 18:

If 3 sin xy+4 cos xy=5, then dydx=

(a) -yx

(b) 3 sin xy+4 cos xy3 cos xy-4 sin xy

(c) 3 cos xy+4 sin xy4 cos xy-3 sin xy

(d) none of these

Answer:

(a) -yx

We have, 3 sinxy+4 cosxy=5     3 cosxyxdydx+y-4 sinxyxdydx+y=0   xdydx+y3 cosxy-4 sinxy =0   xdydx+y=0   xdydx=-y     dydx=-yx

Page No 10.119:

Question 19:

If sin y=x sin a+y, then dydx is

(a) sin asin a sin2 a+y

(b) sin2 a+ysin a

(c) sin a sin2 a+y

(d) sin2 a-ysin a

Answer:

(b) sin2 a+ysin a

We have, sin y=x sina+y

ddxsin y=ddxx sina+ycos ydydx=sina+yddxx+xddxsina+ycos ydydx=sina+y×1+x cosa+ydydx cos ydydx=sina+y+x cosa+ydydxcos ydydx-x cosa+ydydx=sina+ycos y-x cos a+ydydx=sina+ycos y-sin ysina+y×cosa+ydydx=sina+y                       sin y=2 sinx cosxx=sin ysina+ysina+y cos y-sin y cosa+ysina+ydydx=sina+ysina+y-ysina+y×dydx=sina+y dydx=sin2a+ysina   

Page No 10.119:

Question 20:

The derivative of cos-1 2x2-1 with respect to cos-1 x is
(a) 2

(b) 12 1-x2

(c) 2/x

(d) 1-x2

Answer:

(a) 2

Let u=cos-12x2-1Put x=cosθθ=cos-1xdθdx=-11-x2Now, u=cos-1cos2θu=2θ

dudx=2dθdxdudx=-21-x2               ...iand, v=cos-1xv=cos-1cosθv=θ

dvdx=dθdxdvdx=-11-x2                       ...iiDividing i by ii, we get,dudxdvdx=-21-x2×1-x2-1dudv=2

Page No 10.119:

Question 21:

If fx=x2+6x+9, then f'x is equal to
(a) 1 for x<-3
(b) -1 for x<-3
(c) 1 for all xR
(d) none of these

Answer:

b-1  for x<-3 We have, fx=x2+6x+9                        = x+32                           =x+3 fx=x+3,       x-3 -x-3,      x<-3   f'x=1,       x-3 -1,      x<-3   f'x=-1  for x<-3  

Page No 10.119:

Question 22:

If fx=x2-9x+20, then f' (x) is equal to
(a) -2x+9 for all xR
(b) 2x-9 if 4<x<5
(c) -2x+9, if 4<x<5
(d) none of these

Answer:

c-2x+9 for 4<x<5 We have, fx=x2-9x+20 fx=x2-9x+20,       -<x4-x2-9x+20,       4<x<5x2-9x+20,       5x<fx=2x-9,       -<x4-2x+9,       4<x<52x-9,       5x<f'x=-2x+9 for 4<x<5    

Page No 10.119:

Question 23:

If fx=x2-10x+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, fx=x2-10x+25=x-52=x-5 =x-5  for x>5 -x-5 for x<5LHD=limx5- fx-fax-a        =limx5-x2-10x+25-52-105+25x-5        =limx5-x-5x-5        =limx5--x-5x-5        =-1RHD=limx5+ fx-fax-a        =limx5+x2-10x+25-52-105+25x-5        =limx5+x-5x-5        =limx5+x-5x-5        =1Here, LHDRHDThus, the functionis not differentiable at x=5

Page No 10.119:

Question 24:

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

Answer:

(c) 0

For, x>10fx=x-3=x-3gx=fof x=x-3-3                       =x-3-3                       =x-6g'x =1

Page No 10.119:

Question 25:

If fx=xlxml+m xmxnm+n xnxln+1, the f' (x) is equal to
(a) 1
(b) 0
(c) xl+m+n
(d) none of these

Answer:

(b) 0
We have, fx=xlxml+m xmxnm+n xnxln+1

fx=xl-ml+m × xm-nm+n × xn-ln+lfx=xl2-m2 × xm2-n2 × xn2-l2fx=xl2-m2+m2-n2+n2-l2fx=x0fx=1f'x=0    

Page No 10.119:

Question 26:

If y=11+xa-b+c-b+11+xb-c+xa-c+11+xb-a+xc-a, then dydx is equal to
(a) 1
(b) a+b+cxa+b+c-1
(c) 0
(d) none of these

Answer:

(c) 0


We have,y=11+xa-b+xc-b+11+xb-c+xa-c+11+xb-a+xc-a=11+xaxb+xcxb+11+xbxc+xaxc+11+xbxa+xcxa=xbxb+xa+xc+xcxc+xb+xa+xaxa+xb+xc=xbxa+xb+xc+xcxa+xb+xc+xaxa+xb+xc=xb+xc+xaxa+xb+xc=xa+xb+xcxa+xb+xc=1dydx=ddx1=0



Page No 10.120:

Question 27:

If 1-x6+1-y6=a3 x3-y3, then dydx is equal to
(a) x2y2 1-y61-x6

(b) y2x21-y61+x6

(c) x2y21-x61-y6

(d) none of these

Answer:

(a) x2y2 1-y61-x6

We have,   1-x6+1-y6=ax3-y3Putting x3=sinA and y3=sinB1-sin2A+1-sin2B=asinA-sinB cosA+cosB=asinA-sinB2cosA+B2cosA-B2=2a sinA-B2cosA+B2cotA-B2=aA-B2=cot-1a A-B=2 cot-1asin-1x3-sin-1y3=2 cot-1a

11-x6×ddxx3-11-y6×ddxy3=011-x6×3x2-11-y6×3y2×dydx=0dydx=x2y21-y61-x6

Page No 10.120:

Question 28:

If y=log tan x, then the value of dydxat x=π4 is given by
(a) ∞
(b) 1
(c) 0
(d) 12

Answer:

(b) 1

We have,  y=logtanxdydx=1tanx×ddxtanxdydx=1tanx×12tanx×ddxtanxdydx=sec2x2 tanxNow, dydxx=π4=secπ422 tanπ4=22×1=1

Page No 10.120:

Question 29:

If sin-1 x2-y2x2+y2=log a then dydx is equal to

(a) x2-y2x2+y2

(b) yx

(c) xy

(d) none of these

Answer:

(b) yx
We have, sin-1x2-y2x2+y2=loga  x2-y2x2+y2=sin loga

x2+y22x-2ydydx-x2-y22x+2ydydxx2+y22=02x3-2x2ydydx+2xy2-2y3dydx-2x3-2x2ydydx+2xy2+2y3dydxx2+y22=0-4x2ydydx+4xy2=0-4x2ydydx=-4xy2dydx=4xy24x2ydydx=yx

Page No 10.120:

Question 30:

If sin y=x cos a+y, then dydx is equal to
(a) cos2 a+ycos a

(b) cos acos2 a+y

(c) sin2 ycos a

(d) none of these

Answer:

(a) cos2 a+ycos a

We have, sin y=x cosa+y

ddxsin y=ddxx cosa+ycos ydydx=1×cosa+y-x sina+yddxa+ycos ydydx=cosa+y-x sina+ydydxcos ydydx+x sina+ydydx=cosa+ycos y+x sina+ydydx=cosa+ycos y+sin ycosa+y×sina+ydydx=cosa+y                         sin y=x cosa+yx=sin ycosa+ycosa+y cos y+sin y sina+ycosa+ydydx=cosa+ycosa+y-ycosa+y×dydx=cosa+ydydx=cos2a+ycos a

Page No 10.120:

Question 31:

If y=log 1-x21+x2, then dydx=

(a) 4x31-x4

(b) -4x1-x4

(c) 14-x4

(d) -4x31-x4

Answer:

(b) -4x1-x4
We have, y=log1-x21+x2dydx=11-x21+x2ddx1-x21+x2dydx=1+x21-x21+x2-2x-1-x22x1+x22dydx=11-x2-2x-2x3-2x+2x31+x2dydx=-4x1-x4

Page No 10.120:

Question 32:

If y=tan-1 sin x+cos xcos x-sin x, then dydx is equal to
(a) 12
(b) 0
(c) 1
(d) none of these

Answer:

(c) 1

We have, y=tan-1sinx+cosxcosx-sinxdydx=11+sinx+cosxcosx-sinx2ddxsinx+cosxcosx-sinxdydx=cosx-sinx2cosx-sinx2+sinx+cosx2cosx-sinxddxsinx+cosx-sinx+cosxddxcosx-sinxcosx-sinx2dydx=cosx-sinx2cosx-sinx2+sinx+cosx2cosx-sinxcosx-sinx-sinx+cosx-sinx-cosxcosx-sinx2dydx=cosx-sinx2cosx-sinx2+sinx+cosx2cosx-sinxcosx-sinx+sinx+cosxsinx+cosxcosx-sinx2dydx=cosx-sinx2cosx-sinx2+sinx+cosx2×cosx-sinx2+sinx+cosx2cosx-sinx2dydx=1

Page No 10.120:

Question 33:

If sec11+x1-y=a, then dydx is equal to
(ax-1y-1       (b) x-1y+1        (c) y-1x+1       (d) y+1x-1

Answer:

Given: sec-11+x1-y=a1+x1-y=sec(a)
Differentiating w. r. t. x, we get
1-y1-1+x-dydx1-y2=0
1-y=-1+xdydxdydx=1-y1+x=y-1x+1
Hence, the correct answer is option C.

Page No 10.120:

Question 1:

If y = x x, then dydxx=-1=______________________.

Answer:


We know

x=x,x0-x,x<0

y=xx=x2,x0-x2,x<0

dydx=2x,x0-2x,x<0

For x < 0,

dydx=-2x

dydxx=-1=-2×-1=2


If y = xx, then dydxx=-1=       2      .

Page No 10.120:

Question 2:

If y = 2xx, then dydxx=-1=_________________ and dydxx=1=____________________.

Answer:


We know

x=x,x0-x,x<0

y=2x+x=2x+x,x02x-x,x<0

y=2x+x=3x,x0x,x<0

For x ≥ 0,

y = 3x

dydx=3

dydxx=1=3

For x < 0,

y = x

dydx=1

dydxx=-1=1

Thus, if y = 2x + |x|, then dydxx=-1=1 and dydxx=1=3.


If y = 2xx, then dydxx=-1=     1      and dydxx=1=     3     .

Page No 10.120:

Question 3:

If f(x) = x2-x, then f'(2)=_________________.

Answer:


fx=x2-x=x2-x,x2-x0-x2-x,x2-x<0

fx=x2-x=x2-x,xx-10x-x2,xx-1<0

fx=x2-x=x2-x,-<x0 or 1x<x-x2,0<x<1

For x ≥ 1,

fx=x2-x

f'x=2x-1

f'2=2×2-1=3



If f(x) = x2-x, then f'(2)=      3      .



Page No 10.121:

Question 4:

If y = sinxo and dydx = k cos xo , then k = ________________.

Answer:


y=sinx°

y=sinπ180x                      180°=π radians

Differentiating both sides with respect to x, we get

dydx=ddxsinπ180x

dydx=cosπ180x×ddxπ180x

dydx=cosx°×π180

dydx=π180cosx°

Comparing with dydx=kcosx°, we get

k=π180


If y = sinxo and dydx = k cosxo , then k =       π180       .

Page No 10.121:

Question 5:

If f(x) = exg(x), g(0) = 2, g'(0) = 1, then f'(0) = __________________.

Answer:


fx=exgx

Differentiating both sides with respect to x, we get

ddxfx=ddxexgx

f'x=ex×ddxgx+gx×ddxex

f'x=ex×g'x+gx×ex

Putting x = 0, we get

f'0=e0×g'0+g0×e0

f'0=1×1+2×1                 [g(0) = 2, g'(0) = 1 and e0 = 1]

f'0=1+2=3

Thus, the value of f'0 is 3.


If f(x) = exg(x), g(0) = 2, g'(0) = 1, then f'(0) = ____3____.

Page No 10.121:

Question 6:

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

Answer:


We know

x+2=x+2,x-2-x+2,x<-2

fx=3x+2=3x+2,x-2-3x+2,x<-2

f'x=3,x-2-3,x<-2

For x < −2, f'x=-3

f'-3=-3



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

Page No 10.121:

Question 7:

If f(1) = 3, f'(2) = 1, then ddxIn fex+2x= ____________________________.

Answer:


Disclaimer: The solution is provided for the following question.

If f(1) = 3, f'(1) = 1, then ddxIn fex+2x= ____________________________.


Solution:

ddxlnfex+2x=1fex+2x×ddxfex+2x=1fex+2x×f'ex+2xddxex+2x=1fex+2x×f'ex+2x×ex+2

Putting x = 0, we get

ddxlnfex+2xx=0

=1fe0+2×0×f'e0+2×0×e0+2

=1f1×f'1×3            e0=1

=13×1×3                         f1=3, f'1=1

= 1


If f(1) = 3, f'(1) = 1, then ddxIn fex+2x=      1      .

Page No 10.121:

Question 8:

If f(x) = x x, then ​f '(x) = _________________.

Answer:


We know

x=x,x0-x,x<0

fx=xx=x2,x0-x2,x<0

f'x=2x,x0-2x,x<0

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


If f(x) = xx, 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=x-1,x1-x-1,x<1

x-3=x-3,x3-x-3,x<3

fx=x-1+x-3=-x-1-x-3,x<1x-1-x-3,1x<3x-1+x-3,x3

fx=x-1+x-3=-2x+4,x<12,1x<32x-4,x3

For 1 ≤ x < 3, f(x)  = 2

f'x=0, for 1 ≤ x < 3

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) = cosx-sinx, then f'π3=___________________.

Answer:


For π4<xπ2,

cosx<sinx

cosx-sinx<0

fx=cosx-sinx=-cosx-sinx

fx=-cosx+sinx

Differentiating both sides with respect to x, we get

ddxfx=ddx-cosx+ddxsinx

f'x=--sinx+cosx

f'x=sinx+cosx

f'π3=sinπ3+cosπ3=32+12=3+12         π4<π3π2


​If f(x) = cosx-sinx, then f'π3=           3+12          .

Page No 10.121:

Question 11:

​If f(x) = cos x, then f'π4 =______________________.

Answer:


For 0x<π2,

cosx > 0

fx=cosx=cosx

f'x=-sinx

f'π4=-sinπ4=-12


​If f(x) = cos x, then f'π4 =           -12            .

Page No 10.121:

Question 12:

The derivative of x2 with respect to x3 is __________________.

Answer:


Let u(x) = x2 and v(x) = x3.

ux=x2

dudx=2x

vx=x3

dvdx=3x2

dudv=dudxdvdx

dudv=2x3x2=23x

Thus, the derivative of x2 with respect to x3 is 23x.


The derivative of x2 with respect to x3 is      23x     .












 

Page No 10.121:

Question 13:

For the curve x+y=1, dydx at 14,14 is ______________________.

Answer:


x+y=1         (Given)

Differentiating both sides with respect to x, we get

ddxx+ddxy=ddx1

12x+12ydydx=0

12ydydx=-12x

dydx=-yx=-yx

dydxπ4,π4=-1414

dydxπ4,π4=-1

Thus, the value of dydx at π4,π4 is −1.


For the curve x+y=1, dydx at 14,14 is      -1        .

Page No 10.121:

Question 14:

​If f(x) = sin x, then f'-π4 = ____________________.

Answer:


For -π2<x<0,

sinx<0

fx=sinx=-sinx

f'x=-cosx

f'-π4=-cos-π4=-cosπ4=-12                 cos-θ=cosθ


If f(x) = sin x, then f'-π4 =              -12              .

Page No 10.121:

Question 15:

​If f(x) = sin x - cosx, then f'π6 = ________________________.

Answer:


For 0<x<π4,

sinx<cosx

sinx-cosx<0

fx=sinx-cosx=-sinx-cosx

fx=-sinx+cosx

Differentiating both sides with respect to x, we get

f'x=-cosx-sinx

f'π6=-cosπ6-sinπ6=-32-12=-123+1


If f(x) = sin x - cosx, then f'π6 =                -123+1                 .

Page No 10.121:

Question 16:

If y = tan xo, then dydxx=45o = ________________________.

Answer:


y=tanx°

y=tanπ180x           180°=π radians

Differentiating both sides with respect to x, we get

dydx=ddxtanπ180x

dydx=sec2π180x×ddxπ180x

dydx=π180sec2π180x

dydx=π180sec2x°

dydxx=45°=π180sec245°

dydxx=45°=π180×22=π90


If y = tanxº, then dydxx=45o =      π90     .

Page No 10.121:

Question 17:

If y = sin-1(ex) + cos-1(ex), then dydx = ____________________.

Answer:


y=sin-1ex+cos-1ex

y=π2          sin-1x+cos-1x=π2

Differentiating both sides with respect to x, we get

dydx=ddxπ2

dydx=0


If y = sin−1(ex) + cos−1(ex), then dydx = ____0____.

Page No 10.121:

Question 18:

If y = sin-1(3x-4x3), 12< 1, then dydx = ______________________.

Answer:


y = sin−1(3x − 4x3)

Let x = sinθ.

y=sin-13sinθ-4sin3θ

y=sin-1sin3θ

Now,

12<x<1

12<sinθ<1

π6<θ<π2

π2<3θ<3π2

y=sin-1sin3θ

y=sin-1sinπ-3θ

y=π-3θ              π2<3θ<3π2-π2<π-3θ<π2

y=π-3sin-1x

Differentiating both sides with respect to x, we get

dydx=0-3×11-x2dydx=-31-x2


If y = sin−1(3x − 4x3), 12< 1, then dydx =       -31-x2      .

Page No 10.121:

Question 19:

If y = sec-1 x+1x-1+sin-1x-1x+1, then dydx is equal to ___________________.

Answer:


We know

sec-1a=cos-11a

sec-1x+1x-1=cos-1x-1x+1      .....(1)

Now,

y=sec-1x+1x-1+sin-1x-1x+1

y=cos-1x-1x+1+sin-1x-1x+1        [Using (1)]

y=π2            sin-1a+cos-1a=π2

Differentiating both sides with respect to x, we get

dydx=0

Thus, if y=sec-1x+1x-1+sin-1x-1x+1, then dydx=0.


If y=sec-1x+1x-1+sin-1x-1x+1 , then dydx 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.

ux=cosx

dudx=ddxcosx

dudx=-sinx     .....(1)

vx=sinx

dvdx=ddxsinx

dvdx=cosx        .....(2)

dudv=dudxdvdx

dudv=-sinxcosx       [From (1) and (2)]

dudv=-tanx

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=log10x.

y=log10x

y=logexloge10            logba=logcalogcb

Differentiating both sides with respect to x, we get

dydx=ddxlogexloge10

dydx=1loge10×ddxlogex

dydx=1loge10×1x

dydx=1xloge10

Thus, the derivative of log10x with respect to x is 1xloge10.

The derivative of log10x with respect to x is       1xloge10      .

Page No 10.121:

Question 22:

Ifddx(f(x))=11+x2, then ddxfx3=_________________________.

Answer:


ddxfx3=f'x3×ddxx3

ddxfx3=f'x3×3x2     .....(1)

Now,

ddxfx=f'x=11+x2          (Given)

Replacing x by x3, we get

f'x3=11+x32

f'x3=11+x6         .....(2)

From (1) and (2), we get

ddxfx3=11+x6×3x2

ddxfx3=3x21+x6


Ifddx(f(x))=11+x2, then ddxfx3=     3x21+x6     .

Page No 10.121:

Question 23:

If y = cos (sin x2), then dydx at x=π2 is equal to ______________________.

Answer:


y=cossinx2

Differentiating both sides with respect to x, we get

dydx=ddxcossinx2

dydx=-sinsinx2×ddxsinx2

dydx=-sinsinx2×cosx2×ddxx2

dydx=-sinsinx2×cosx2×2x

dydx=-2xcosx2sinsinx2

Putting x=π2, we get

dydxx=π2=-2×π2×cosπ22sinsinπ22

dydxx=π2=-2×π2×cosπ2×sinsinπ2

dydxx=π2=-2×π2×0×sin1

dydxx=π2=0

Thus, dydx at x=π2 is 0.


If y = cos (sin x2), then dydx at x=π2 is equal to ___0___.

Page No 10.121:

Question 24:

If y = logx, x0, then dydx = ___________________.

Answer:


For y=logx to be defined,

x0

x0

Now,

y=logx=logx,x>0log-x,x<0

dydx=1x,x>01-x×-1,x<0

dydx=1x,x>01x,x<0

dydx=1x, x0


If y = logx, x0, then dydx =        1x       .

Page No 10.121:

Question 25:

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

Answer:


fx=ax2+bx+c

f'x=ddxax2+bx+c

f'x=2ax+b

f'1+f'4-f'5

=2a×1+b+2a×4+b-2a×5+b

=2a+b+8a+b-10a-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) − '(5) is equal to ___b___.

Page No 10.121:

Question 26:

If '(1) = 2 and g'2 = 4, then the derivative of f(tan x) with respect of g(secx) at xπ4 is equal to ______________.

Answer:


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

ux=ftanx

dudx=ddxftanx

dudx=f'tanxddxtanx

dudx=f'tanx×sec2x          .....(1)

vx=gsecx

dvdx=ddxgsecx

dvdx=g'secxddxsecx

dvdx=g'secx×secxtanx           .....(2)

dudv=dudxdvdx

dudv=sec2x×f'tanxsecxtanx×g'secx

dudv=secx×f'tanxtanx×g'secx

Putting x=π4, we get

dudvx=π4=secπ4×f'tanπ4tanπ4×g'secπ4

dudvx=π4=2×f'11×g'2

dudvx=π4=2×21×4

dudvx=π4=12

Thus, the derivative of f(tan x) with respect of g(secx) at x = π4 is 12.


If '(1) = 2 and g'2 = 4, then the derivative of f(tan x) with respect of g(secx) at x = π4 is equal to      12     .



Page No 10.122:

Question 1:

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

Answer:

We have,  fx=loge logex
Differentiating with respect to x,
f'x=1logexddxlogex   f'x=1logex1xf'e=1logee1e               x=ef'e=1e                  loge e=1

Page No 10.122:

Question 2:

If fx=x+1, then write the value of ddx fof x.

Answer:

We have,  fx=x+1 Now, fofx=ffx  fofx=fx+1fofx=x+1+1fof=x+2

ddxfofx=ddxx+ddx2ddxfofx=1+0ddxfofx=1

Page No 10.122:

Question 3:

If f'1=2 and y=f loge x, find dydx at x=e.

Answer:

We have,  f'1=2 and y=flogex
Differentiate it with respect to x,
dydx=f'logex×ddxlogexdydx=f'logex1xdydx=f'logee1e         x=edydx=f'11e                  logee=1dydx=2e                            f'1=2

Page No 10.122:

Question 4:

If f1=4, f'1=2, find the value of the derivative of log fex w.r. to x at the point x = 0.

Answer:

We have, f1=4 and f'1=2Let y=logfex

dydx=ddxlogfexdydx=1fex×ddxfexdydx=1fex×f'ex×ddxexdydx=exf'exfexPutting x=0, we get,dydx=e0f'e0fe0dydx=1f'1f1dydx=24                             f'1=2 and f1=4dydx=12

Page No 10.122:

Question 5:

If f'x=2x2-1 and y=f x2, then find dydx at x=1.

Answer:

We have, f'x=2x2-1and  y=fx2

dydx=ddxfx2dydx=f'x2ddxx2dydx=f'x2× 2xdydx=2xf'x2Putting x=1, we get,dydx=21f'12dydx=2 ×f'1dydx=2 ×1                               f'1=212-1=2-1=1dydx=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, f3=9 ,f'3=9and gx=f-1xgofx=xgfx=x

ddxgfx=1g'fxddxfx=1g'fx × f'x=1Puting x=3, we get,g'f3 × f'3=1g'9 × 9=1                               f3=9 , f'3=9g'9=19

Page No 10.122:

Question 7:

If y=sin-1 sin x,-π2xπ2. Then, write the value of dydx for x  -π2, π2.

Answer:

We have, y=sin-1sinxy=x                          sin-1sinx=x , if x-π2,π2

dydx=ddxxdydx=1

Page No 10.122:

Question 8:

If π2x3π2 and y=sin-1 sin x, find dydx.

Answer:

We have, y=sin-1sinx                               y=π-x                        sin-1sinx=π-x , if xπ2,3π2                          

dydx=ddxπ-xdydx=0-1dydx=-1

Page No 10.122:

Question 9:

If πx2π and y=cos-1 cos x, find dydx.

Answer:

We have, y=cos-1cosx   y=2π-x                               cos-1cosx=2π-x , if xπ,2π                       

dydx=ddx2π-xdydx=0-1dydx=-1

Page No 10.122:

Question 10:

If y=sin-1 2x1+x2, write the value of dydxfor x>1.

Answer:

We have, y=sin-12x1+x2Putting x=tanθ1<tanθ<π4<θ<π2    π2<2θ<πy=sin-1sin2θ  y=sin-1sinπ-2θ   y=π-2θ  y=π-2tan-1xdydx=0-21+x2  dydx=-21+x2                        
 

Page No 10.122:

Question 11:

If f0=f1=0, f'1=2 and y=f ex ef x, write the value of dydx at x=0.

Answer:

We have, f0=f1=0 , f'1=2and,y=fexefx

dydx=ddxfex × efxdydx=fexddxefx+efxddxfex                   Using product ruledydx=fex × efxddxfx+efx×f'exddxexdydx=fex × efx×f'x+efx×f'ex×exPutting x=0, we get,dydx=fe0 × ef0×f'0+ef0×f'e0×e0dydx=f1ef0×f'0+ef0×f'1×1dydx=0×e0×f'0+e0×2×1                             fx=f1=0 and f'1=2dydx=0+1×2×1dydx=2

Page No 10.122:

Question 12:

If y=x x, find dydx for x<0.

Answer:

We have, y=xxy=x-x  x<0y=-x2

dydx=ddx-x2dydx=-2x

Page No 10.122:

Question 13:

If y=sin-1 x+cos-1 x, find dydx.

Answer:

We have, y=sin-1x+cos-1x y=π2                             sin-1x+cos-1x=π2         

dydx=0

Page No 10.122:

Question 14:

If x=a θ+sin θ, y=a 1+cos θ, find dydx.

Answer:

We have, x=aθ+sinθ   and y=a1+cosθdxdθ=addθθ+ddθsinθ and dydθ=a0-sinθdxdθ=a1+cosθ and dydθ=-asinθ dydx=dydθdxdθ=-asinθa1+cosθ=-2sinθ2cosθ22 cos2θ2=-tanθ2          
 

Page No 10.122:

Question 15:

If -π2<x<0 and y=tan-1 1-cos 2x1+cos 2x, find dydx.

Answer:

We have, y=tan-11-cos2x1+cos2xy=tan-12 sin2x2 cos2xy=tan-1tan2xy=tan-1tanx                         tan-1tanx=-x , if x-π2,0y=-x

dydx=-1

Page No 10.122:

Question 16:

If y=xx, find dydx at x=e.

Answer:

We have, y=xx     ...i
Taking log on both sides,
log y=log xxlog y=x logx

1ydydx=xddxlogx+logxddxx1ydydx=x1x+logx 11ydydx=1+logxdydx=y1+logxdydx=xx1+logx                                using equation iPuting x=e, we get,dydx=ee1+logeedydx=ee1+1                                      logee=1dydx=2ee 



Page No 10.123:

Question 17:

If y=tan-1 1-x1+x, find dydx.

Answer:

We have, y=tan-11-x1+x

dydx=11+1-x1+x2ddx1-x1+xdydx=1+x21+x2+2x+1+x2-2x1+xddx1-x-1-xddx1+x1+x2                  using quotient ruledydx=1+x22x2+21+x-1-1-x11+xdydx=1+x22x2+1-x-1-1+x1+x2dydx=1+x22x2+1×-21+x2dydx=-1x2+1

Page No 10.123:

Question 18:

If y=loga x, find dydx.

Answer:

We have, y=logaxy=logxloga                      logab=logbloga

dydx=1logaddxlogxdydx=1loga1xdydx=1x loga

Page No 10.123:

Question 19:

If y=log tan x, write dydx.

Answer:

We have, y=logtanxy=logtanx12y=12log tanx                      logab=bloga

dydx=12×1tanxddxtanxdydx=12×1tanxsec2xdydx=12sinxcosx×cos2xdydx=12 sinx cosxdydx=1sin2xdydx=cosec2x

Page No 10.123:

Question 20:

If y=sin-1 1-x21+x2+cos-1 1-x21+x2, find dydx.

Answer:

We have, y=sin-1 1-x21+x2+cos-1 1-x21+x2y=π2       sin-1x+cos-1x=π2   

dydx=0

Page No 10.123:

Question 21:

If y=sec-1 x+1x-1+sin-1 x-1x+1, then write the value of dydx.

Answer:

We have, y=sec-1x+1x-1+sin-1x-1x+1                    y=cos-1x-1x+1+sin-1x-1x+1                   sec-1x=cos-11xy=π2      sin-1x+cos-1x=π2

dydx=0

Page No 10.123:

Question 22:

If x <1 and y=1+x+x2+... to ∞, then find the value of dydx.

Answer:

We have, y=1+x+x2+....to  y=11-x                            It is a G.P with first term  1and common ratio x

dydx=ddx11-xdydx=-11-x2ddx1-xdydx=-11-x2-1dydx=11-x2

Page No 10.123:

Question 23:

If u=sin-1 2x1+x2 and v=tan-1 2x1-x2, where -1<x<1, then write the value of dudv.

Answer:

We have, u=sin-12x1+x2 and v=tan-12x1-x2dudx=21+x2 and dvdx=21+x2              dudv=dudxdvdx=21+x2×1+x22=1

Page No 10.123:

Question 24:

If fx=log u xv x, u 1=v 1 and u' 1=v' 1=2, then find the value of f' (1).

Answer:

We have, fx=loguxvxand, u1=v1 , u'1=v'1=2            ...i

f'x=ddxloguxvxf'x=1uxvx×ddxuxvxf'x=vxux×vxddxux-uxddxvxvx2                     f'x=vxux×vx×u'x-ux×v'xvx2Putting x=1, we get,f'1=v1u1×v1×u'1-u1×v'1v12f'1=1×u1×2-u1×2u12      Using eqn 1f'1=0u12             f'1=0

Page No 10.123:

Question 25:

If y=log 3x, x0, find dydx.

Answer:

We have, y=log3x
dydx=ddxlog3xdydx=13xddx3xdydx=13x3dydx=1x

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, fx is an even function.f-x=fx

ddxf-x=ddxfxf'-xddx-x=f'xf'-x×-1=f'x-f'-x=f'xf'-x=-f'xThus, 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, fx is an odd function.f-x=-fx

ddxf-x=-ddxfxf'-xddx-x=-f'xf'-x×-1=-f'x-f'-x=-f'xf'-x=f'xThus, f'x is an even function.

Page No 10.123:

Question 28:

If x=3sint-sin3t, y=3cost-cos3t finddydx at t=π3

Answer:

x=3sint-sin3t and y=3cost-cos3tdxdt=3cost-3cos3t and dydt=-3sint+3sin3tdydx=dydtdxdt=-3sint+3sin3t3cost-3cos3tdydxt=π3=-3sinπ3+3sinπ3cosπ3-3cosπ=-3×32+03×12+3=-33292=-13

Page No 10.123:

Question 29:

If y = log (cos ex), then find dydx.

Answer:


y=logcosex

Differentiating both sides with respect to x, we get

dydx=ddxlogcosex

dydx=1cosex×ddxcosex

dydx=1cosex×-sinex×ddxex

dydx=-sinexcosex×ex

dydx=-extanex

 

Page No 10.123:

Question 30:

If f(x) = x + 7 and g(x) = x – 7, x ∈ R, then find ddxfogx.

Answer:


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

fogx

=fgx

=gx+7                  fx=x+7

=x-7+7                 gx=x-7

=x

ddxfogx=ddxx

ddxfogx=1

Thus, the value of ddxfogx is 1.


 



Page No 10.17:

Question 1:

Differentiate the following functions from first principles:

ex

Answer:

Let  fx=e-xfx+h=e-x+h ddxfx=limh0fx+h-fxh                  =limh0e-x+h-e-xh                  =limh0e-x×e-h-e-xh                  =limh0e-xe-h-1-h×-1                  =-e-x limh0e-h-1-h                                                         =-e-x                    limh0e-h-1-h=1So, ddxe-x=-e-x

Page No 10.17:

Question 2:

Differentiate the following functions from first principles:

e3x

Answer:

Let  fx=e3xfx+h=e3x+hddxfx=limh0fx+h-fxh                =limh0e3x+h-e3xh                =limh0e3xe3h-e3xh                =limh0e3xe3h-13h×3                =3e3xlimh0e3h-13h                =3e3xHence, ddxe3x=3e3x

Page No 10.17:

Question 3:

Differentiate the following functions from first principles:

eax+b

Answer:

Let  fx=eax+b       fx+h=eax+h+b        ddxfx=limh0fx+h-fxh                             =limh0eax+h+b-eax+bh                             =limh0eax+beah-eax+bh                             =limh0eax+beah-1ah×a                             =aeax+blimh0eah-1ah                             =aeax+bSo, ddxeax+b=aeax+b

Page No 10.17:

Question 4:

Differentiate the following functions from first principles:

ecos x

Answer:

Let  fx=ecosx       fx+h=ecosx+h        ddxfx=limh0fx+h-fxh                             =limh0ecosx+h-ecosxh                             =limh0 ecosxecosx+h-cosx-1h                             =limh0 ecosxecosx+h-cosx-1cosx+h cosx×cosx+h-cosxh                             =ecosxlim             h0  cosx+h-cosxh ×limh0ecosx+h-cosx-1cosx+h-cosx                             =ecosxlimh0  cosx+h-cosxh                            limh0ex-1x=1                             =ecosxlimh0 -2sinx+h+x2sinx+h-x2h          cosA-cosB=-2sin A+B2sinA-B2                            =ecosxlimh0-sin2x+h21×sinh2h2                            =ecosxlimh0-sin2x+h21× lim       h0sinh2h2                           =ecosxlimh0-sin2x+h2                                         sinxx=1                          =ecosx-sinx                           =-sinxecosxHence, ddxecosx=-sinxecosx

Page No 10.17:

Question 5:

Differentiate the following functions from first principles:

e2x

Answer:

Let  fx=e2x  fx+h=e2x+h        ddxfx=limh0fx+h-fxh                             =limh0e2x+h-e2xh                             =limh0 e2xe2x+h-2x-1h                             =e2xlimh0 e2x+h-2x-12x+h-2x×limh02x+h-2xh                             =e2xlimh02x+h-2xh                                 limh0eh-1h=1                             =e2xlimh0 2x+h-2xh × 2x+h+2x2x+h+2x       Rationalising the numerator                            =e2xlimh02x+h-2xh2x+h+2x                           =e2xlimh02x+2h-2xh2x+h+2x                                                                =e2xlimh02hh2x+h+2x                           =e2xlimh022x+h+2x                          =e2x2xHence, ddxe2x=e2x2x

Page No 10.17:

Question 6:

Differentiate the following functions from first principles:

log cos x

Answer:

Let fx=log cosxfx+h=log cosx+hddxfx=limh0fx+h=fxh                     =limh0log cosx+h-log cosxh                     =limh0logcosx+hcosxh                        logA-logB=logAB                     =limh0log1+cosx+hcosx-1h                     =limh0log1+cosx+h-cosxcosxcosx+h-cosxcosx×limh0cosx+h-cosxcosx                     =1×limh0cosx+h-cosxcosx × h                    limx0log1+xx=1                    =limh0-2sinx+h+x2sinx+h-x2cosx × h                     =-2limh0sin2x+h2×sinh22cosx ×h2                     =-2sinx2cosx                            limx0sinxx=1                      =-tanxSo, ddxlog cosx=-tanx

Page No 10.17:

Question 7:

Differentiate the following function from first principles:

ecot x
 

Answer:

  Let  fx=ecotxfx+h=ecotx+h  ddxfx=limh0fx+h-fxh                      =limh0ecotx+h-ecotxh                      =limh0ecotxecotx+h-cotx-1h                      =ecotxlimh0ecotx+h-cotx-1cotx+h-cotx×cotx+h-cotxh                      =ecotxlimh0cotx+h-cotxh×cotx+h+cotxcotx+h+cotx                            limx0ex-1x=1 and rationalizing the numerator                      =ecotxlimh0cotx+h-cotxhcotx+h+cotx                      =ecotxlimh0cotx+hcotx+1cotx-x-hhcotx+h+cotx                                                                     cotA-B=cotAcotB+1cotB-cotA                      =ecotxlimh0cotx+hcotx+1cot-h×hcotx+h+cotx                      =-ecotxlimh0cotx+hcotx+1htanhcotx+h+cotx                      =ecotx×cot2x+12cotx                                       limx0tanxx=1                      =-ecotx×cosec2x2cotx                                          1+cot2x=cosec2xddxecotx=-ecotx×cosec2x2cotx  

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Question 8:

Differentiate the following functions from first principles:

x
2ex
 

Answer:

 Let  fx=x2exfx+h=x+h2ex+h                 =limh0fx+h-fxh                 =limh0x+h2ex+h-x2exh                 =limh0x2ex+h-x2exh+2xhex+hh+h2ex+hh                 =limh0x2exex+h-x-1h+2xex+h+hex+h                 =limh0x2exeh-1h+2xex+h+hex+h                 =x2ex+2xex+0xex                        limx0ex-1x=1ddxx2ex=exx2+2x

Page No 10.17:

Question 9:

Differentiate the following functions from first principles:

 log cosec x

Answer:

 Let  fx=log cosecx         fx+h=log cosecx+hddxfx=limh0fx+h-fxh                     =limh0log cosecx+h-log cosecxh                     =limh0logcosecx+hcosecxh                     =limh0log1+sinxsinx+h-1h                     =limh0log1+sinx-sinx+hsinx+hsinx-sinx+hsinx+hsinx-sinx+hsinx+hh                     =limh02cosx+x+h2sinx-x-h2sinx+hh                             limx0log1+xx=1 and sinA-sinB=2cosA+B2sinA-B2                     =limh02cos2x+h2sinx+h -2sin-h2-h2                                 limx0sinxx=1                     =-cotxddxlog cosecx=-cotx

Page No 10.17:

Question 10:

Differentiate the following functions from first principles:

sin−1 (2x + 3)

Answer:

 Let fx=sin-12x+3        fx+h=sin-12x+h+3        fx+h=sin-12x+2h+3 ddxfx=limh0fx+h-fxh                      =limh0sin-12x+2h+3-sin-12x+3h                      =limh0sin-12x+2h+31-2x+32-2x+31-2x+2h+32h             sin-1x-sin-1y=sin-1x1-y2-y1-x2                      =limh0sin-1zz×zhwhere, z=2x+2h+31-2x+32-2x+31-2x+2h+32 and limh0sin-1hh=1                      =limh0zh                      =limh0 2x+2h+31-2x+32-2x+31-2x+2h+32 h                      =limh02x+2h+321-2x+32-2x+321-2x+2h+32h2x+2h+31-2x+32+2x+31-2x+2h+32           Rationalizing numerator                      =limh02x+32+4h2+4h2x+31-2x+32-2x+321-2x+32-4h2-4h2x+3h2x+2h+31-2x+32+2x+31-2x+2h+32                      =limh02x+32+4h2+4h2x+3-2x+34-4h22x+32-4h2x+33-2x+32+2x+34+4h22x+32+4h2x+33h2x+2h+31-2x+32+2x+31-2x+2h+32                      =limh04hh+2x+3h2x+2h+31-2x+32+2x+31-2x+2h+32                      =42x+32x+31-2x+32+2x+31-2x+32                      =42x+322x+31-2x+32                      =21-2x+32 ddxsin-12x+3=21-2x+32             



Page No 10.37:

Question 1:

Differentiate

sin (3x + 5)

Answer:

Let y=sin3x+5Differentiating y with respect to x we get,dydx=ddxsin3x+5        =cos3x+5ddx3x+5           using chain rule        =cos3x+5 ×3        =3cos3x+5So,ddxsin3x+5=3cos3x+5

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Question 2:

Differentiate

tan2 x

Answer:

Let  y=tan2xDifferentiating with respect to x we get,dydx=2 tanxddxtanx             using chain rule        =2 tanx × sec2x           So, ddxtan2x=2 tanxsec2x 

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Question 3:

Differentiate

tan (x° + 45°)

Answer:

Let, y=tanx°+45°y=tanx+45π180Differentiating it with respect to x we get,dydx=ddxtanx+45π180        =sec2x+45π180×ddxx+45π180         Using chain rule        =π180sec2x°+45°So,      ddxtanx°+45°=π180sec2x°+45°

Page No 10.37:

Question 4:

Differentiate

sin (log x)

Answer:

Let  y=sinlogxDifferentiate it with respect to x we get,dydx=ddxsinlogx         =coslogxddxlogx        using chain rule         =1xcoslogxSo, ddxsinlogx=1xcoslogx

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Question 5:

Differentiate

esin x

Answer:

Let  y=esinxDifferentiate it with respect to x we get,dydx=ddxesinx        =esinxddxsinx              using chain rule        =esinx×cosxddxx       using chain rule        =esinx×cosx×12x        =cosx esinx2xSo, ddxesinx=cosx esinx2x

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Question 6:

Differentiate

etan x

Answer:

Let y=etanxDifferentiate it with respect to x we get,dydx=ddxetanx        =etanxddxtanx            using chain rule        =etanx × sec2xSo, ddxetanx= sec2xetanx

Page No 10.37:

Question 7:

Differentiate

sin2 (2x + 1)

Answer:

Let  y=sin22x+1Differentiate it with respect to x we get,dydx=ddxsin22x+1        =2sin2x+1ddxsin2x+1                        using chain rule        =2sin2x+1 cos2x+1 ddx2x+1        using chain rule         =4sin2x+1 cos2x+1        =2sin22x+1                                  sin2A=2sinAcosA        =2 sin4x+2So, ddxsin22x+1=2 sin4x+2

Page No 10.37:

Question 8:

Differentiate

log7 (2x − 3)

Answer:

Let y=log72x-3    y=log2x-3log7                      logab=logblogaDifferentiate it with respect to x we get,dydx=1log7ddxlog2x-3        =1log7×12x-3ddx2x-3    using chain rule        =22x-3log7Hence, ddxlog72x-3=22x-3log7

Page No 10.37:

Question 9:

Differentiate

tan 5x°

Answer:

Let y=tan5x°   y=tan5x×π180Differentiate it with respect to x we get,dydx=ddxtan5x×π180         =sec25x×π180ddx5x×π180        using chain rule         =5π180sec25x×π180         =5π180sec25x°Hence,  ddxtan5x°=5π180sec25x°

Page No 10.37:

Question 10:

Differentiate

2x3

Answer:

Let  y=2x3Differentiate it with respect to x we get,dydx=ddx2x3        =2x3×loge2ddxx3            using chain rule        =3x2 × 2x3×loge2Hence, ddx2x3=3x2 × 2x3loge2

Page No 10.37:

Question 11:

Differentiate

3ex

Answer:

Let y=3exDifferentiate it with respect to x we get,dydx=ddx3ex        =3exlog3ddxex           using chain rule        =ex×3exlog3So, ddx3ex=ex×3exlog3

Page No 10.37:

Question 12:

Differentiate

logx 3

Answer:

Let y=logx3   y=log3logx                      logab=logblogaDifferentiate it with respect to x we get,dydx=ddxlog3logx        =log3ddxlogx-1        =log3 × -1logx-2ddxlogx       using chain rule        =-log3logx2×1x        =-log3logx2×1x×1log3        =-1xlog3log3x2                             logbloga=logabSo, ddxlogx3=-1xlog3log3x2

Page No 10.37:

Question 13:

Differentiate

3x2+2x

Answer:

Let y=3x2+2xDifferentiate it with respect to x we get,dydx=ddx3x2+2x        =3x2+2x × loge3ddxx2+2x           using chain rule        =2x+23x2+2x loge3So, ddx3x2+2x=2x+23x2+2x loge3

Page No 10.37:

Question 14:

Differentiate

a2-x2a2+x2

Answer:

Let y=a2-x2a2+x2   y=a2-x2a2+x212Differentiate it with respect to x we get,dydx=ddxa2-x2a2+x212        =12a2-x2a2+x212-1×ddxa2-x2a2+x2                                                       Using chain rule        =12a2-x2a2+x2-12×a2+x2ddxa2-x2-a2-x2ddxa2+x2a2+x22         =12a2+x2a2-x212-2xa2+x2-2xa2-x2a2+x22        =12a2+x2a2-x212-2xa2-2x3-2xa2+2x3a2+x22        =12a2+x2a2-x212-4xa2a2+x22        =-2xa2a2-x2a2+x232So,  ddxa2-x2a2+x2=-2a2xa2-x2a2+x232

Page No 10.37:

Question 15:

Differentiate

3x log x

Answer:

Let  y=3x logxDifferentiate it with respect to x we get,dydx=ddx3x logx        =3x logx×loge3ddxx logx                             Using chain rule        =3x logx×loge3xddxlogx+logxddxx          =3x logx×loge3xx+logx         =3x logx1+logx×loge3So, ddx3x logx=3x logx1+logxloge3

Page No 10.37:

Question 16:

Differentiate

1+sin x1-sin x

Answer:

Let  y=1+sinx1-sinxDifferentiate it with respect to x we get,dydx=ddx1+sinx1-sinx12       =121+sinx1-sinx12-1ddx1+sinx1-sinx       =121-sinx1+sinx121-sinxcosx-1+sinx-cosx1-sinx2       =121-sinx121+sinx12cosx-cosx sinx+cosx+sinx cosx1-sinx2       =12×2cosx1+sinx1-sinx32       =cosx1+sinx1-sinx32       =cosx1+sinx1-sinx1-sinx       =cosx1-sin2x×1-sinx        =cosxcosx1-sinx                        Using 1-sin2x=cos2x        =11-sinx×1+sinx1+sinx        =1+sinx1-sin2x         =1+sinxcos2x         =1cosx1cosx+sinxcosx         =secxsecx+tanxHence, dydx=secxsecx+tanx

Page No 10.37:

Question 17:

Differentiate

1-x21+x2

Answer:

Let  y=1-x21+x2     y=1-x21+x212Differentiate it with respect to x we get,dydx=ddx1-x21+x212        =121-x21+x212-1×ddx1-x21+x2           Using chain rule        =121-x21+x2-12×1+x2ddx1-x2-1-x2ddx1+x21+x22          =121+x21-x212-2x1+x2-2x1-x21+x22         =121+x21-x212-2x-2x3-2x+2x31+x22          =121+x21-x212-4x1+x22           =-2x1-x21+x232

Page No 10.37:

Question 18:

Differentiate

(log sin x)2

Answer:

Let y=log sinx2Differentiate with respect to x we get,dydx=ddxlog sinx2        =2log sinxddxlog sinx        =2log sinx×1sinxddxsinx        =2log sinx×1sinx×cosx        =2log sinxcotxSo, ddxlog sinx2=2log sinxcotx

Page No 10.37:

Question 19:

Differentiate

1+x1-x

Answer:

Let  y=1+x1-x     y=1+x1-x12Differentiate it with respect to x we get,dydx=ddx1+x1-x12        =121+x1-x12-1×ddx1+x1-x           Using chain rule        =121+x1-x-12×1-xddx1+x-1+xddx1-x1-x2   Using quotient rule        =121-x1+x121-x1-1+x-11-x2         =121-x1+x121-x+1+x1-x2          =121-x121+x12×21-x2           =11+x1-x32So, ddx1+x1-x=11+x1-x32

Page No 10.37:

Question 20:

Differentiate

sin 1+x21-x2

Answer:

Let y=sin1+x21-x2Differentiate it with respect to x we get,dydx=ddxsin1+x21-x2         =cosx1+x21-x2ddx1+x21-x2                   Using chain rule         =cosx1+x21-x21-x2ddx1+x2-1+x2ddx1-x21-x22       Using quotient rule         =cosx1+x21-x21-x22x-1+x2-2x1-x22         =cosx1+x21-x22x-2x3+2x+2x31-x22        =4x1-x22cosx1+x21-x2So,ddxsin1+x21-x2=4x1-x22cosx1+x21-x2     

Page No 10.37:

Question 21:

Differentiate

e3 x cos 2x

Answer:

Let  y=e3x cos2xDifferentiate it with respect to x we get,dydx=ddxe3x cos2x        =e3x×ddxcos2x+cos2xddxe3x               Using product rule        =e3x×-sin2xddx2x+cos2xe3xddx3x   Using chain rule        =-2e3x sin2x+3e3x cos2x        =e3x3 cos2x-2 sin2xSo,ddxe3x cos2x=e3x3 cos2x-2 sin2x

Page No 10.37:

Question 22:

Differentiate

sin (log sin x)

Answer:

Let y=sinlog sinxDifferentiate it with respect to x we get,dydx=ddxsinlog sinx        =coslog sinxddxlog sinx                Using chain rule        =coslog sinx×1sinxddx sinx          Using chain rule        =coslog sinxcosxsinx        =coslog sinx cotxHence, ddxsinlog sinx=coslog sinx cotx

Page No 10.37:

Question 23:

Differentiate

etan 3 x

Answer:

Let y=etan3xDifferentiate it with respect to x we get,dydx=ddxetan3x       =etan3xddxtan3x                =etan3xsec23x ×ddx3x       =etan3xsec23x ×3So, ddxetan3x=3etan3xsec23x

Page No 10.37:

Question 24:

Differentiate

ecot x

Answer:

Let  y=ecotx   y=ecotx12Differentiate it with respect to x we get,dydx=ddxecotx12        =ecotx12×ddxcotx12             Using chain rule        =ecotx×12cotx12-1ddxcotx        =-ecotx×cosec2x2cotxSo, ddxecotx=-ecotx×cosec2x2cotx

Page No 10.37:

Question 25:

Differentiate

log sin x1+cos x

Answer:

Let y=logsinx1+cosxDifferentiate it with respect to x, we getdydx=ddxlogsinx1+cosx        =1sinx1+cosx×ddxsinx1+cosx                   Using chain rule        =1+cosxsinx1+cosxddxsinx-sinxddx1+cosx1+cosx2         Using quotient rule        =1+cosxsinx1+cosxcosx-sinx-sinx1+cosx2        =1+cosxsinxcosx+cos2x+sin2x1+cosx2        =1+cosxsinx1+cosx1+cosx2        =1sinx        =cosecxSo, ddxlogsinx1+cosx=cosecx

Page No 10.37:

Question 26:

Differentiate

log 1-cos x1+cos x

Answer:

Let  y=log1-cosx1+cosx   y=log1-cosx1+cosx12   y=12log1-cosx1+cosx                        using logab=blogaDifferentiate it with respect to x we get,dydx=ddx12log1-cosx1+cosx        =12×11-cosx1+cosx×ddx1-cosx1+cosx          Using chain rule         =121+cosx1-cosx1+cosxddx1-cosx-1-cosxddx1+cosx1+cosx2              Using quotient rule         =121+cosx1-cosx1+cosxsinx-1-cosx-sinx1+cosx2         =121+cosx1-cosxsinx+sinx cosx+sinx-sinx cosx1+cosx2         =121+cosx1-cosx2sinx1+cosx2          =sinx1-cosx1+cosx          =sinx1-cos2x          =sinxsin2x                                     =1sinx         =cosecxSo, ddxlog1-cosx1+cosx=cosecx

Page No 10.37:

Question 27:

Differentiate

tan esin x

Answer:

Let y=tanesinxDifferentiate it with respect to x we get, dydx=ddxtanesinx         =sec2esinxddxesinx                Using chain rule         =sec2esinx×esinx×ddxsinx        =cosxsec2esinx×esinxSo, ddxtanesinx=cosxsec2esinxesinx  

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Question 28:

Differentiate

log x+x2+1

Answer:

Let  y=logx+x2+1Differentiate with respect to x we get,dydx=ddxlogx+x2+1        =1x+x2+1ddxx+x2+112                  Using chain rule        =1x+x2+11+12x2+112-1ddxx2+1        =1x+x2+11+12x2+1×2x        =1x+x2+1x2+1+xx2+1        =1x2+1So,  ddxlogx+x2+1=1x2+1

Page No 10.37:

Question 29:

Differentiate

ex log xx2

Answer:

Let   y=exlogxx2Differentiate with respect to x we get,dydx=x2ddxexlogx-exlogxddxx2x22                Using quotient rule        =x2exddxlogx+logxddxex-ex logx×2xx4            Using product rule        =x2exx+ex logx-2xex logxx4        =x2ex1+xlogxx-2xex logxx4        =xex1+xlogx-2logxx4        =xexx31x+x logxx-2logxx        =exx-21x+logx-2xlogxSo, ddxexlogxx2  = exx-21x+logx-2xlogx     

Page No 10.37:

Question 30:

Differentiate

log cosec x-cot x

Answer:

Let y=log cosecx-cotxDifferentiate it with respect to x we get,dydx=ddxlog cosecx-cotx        =1cosecx-cotxddxcosecx-cotx        =1cosecx-cotx×-cosecx cotx+cosec2x        =cosecxcosecx-cotxcosecx-cotx        =cosecxSo,ddxlog cosecx-cotx=cosecx

Page No 10.37:

Question 31:

Differentiate

e2x+e-2xe2x-e-2x

Answer:

Let  y=e2x+e-2xe2x-e-2x
Differentiate with respect to x we get,
dydx=ddxe2x+e-2xe2x-e-2x         =e2x-e-2xddxe2x+e-2x-e2x+e-2xddxe2x-e-2xe2x-e-2x2              Using quotient rule and chain rule         =e2x-e-2xe2xddx2x+e-2xddx-2x-e2x+e-2xe2xddx2x-e-2xddx-2xe2x-e-2x2         =e2x-e-2x2e2x-2e-2x-e2x+e-2x2e2x+2e-2xe2x-e-2x2         =2e2x-e-2x2-2e2x+e-2x2e2x-e-2x2         =2e4x+e-4x-2e2xe-2x-e4x-e-4x-2e2xe-2xe2x-e-2x2         =-8e2x-e-2x2So, ddxe2x+e-2xe2x-e-2x=-8e2x-e-2x2

Page No 10.37:

Question 32:

Differentiate

log x2+x+1x2-x+1

Answer:

Let  y=logx2+x+1x2-x+1

Differentiate with respect of x we get,

dydx=ddxlogx2+x+1x2-x+1        =1x2+x+1x2-x+1ddxx2+x+1x2-x+1             Using chain rule and quotient rule        =x2-x+1x2+x+1x2-x+1ddxx2+x+1-x2+x+1ddxx2-x+1x2-x+12        =x2-x+1x2+x+1x2-x+12x+1-x2+x+12x-1x2-x+12        =x2-x+1x2+x+12x3-2x2+2x+x2-x+1-2x3-2x2-2x+x2+x+1x2-x+12        =-4x2+2x2+2x2+x+1x2-x+1        =-4x2+2x2+2x2+12-x2        =-2x2-1x4+1+2x2-x2        =-2x2-1x4+x2+1So, ddxlogx2+x+1x2-x+1=-2x2-1x4+x2+1

Page No 10.37:

Question 33:

Differentiate

tan-1 ex

Answer:

Let y=tan-1ex
Differentiate it with respect to x we get,
dydx=ddxtan-1ex       =11+ex2ddxex                 Using chain rule       =11+e2x×ex       =ex1+e2xSo, ddxtan-1ex=ex1+e2x

Page No 10.37:

Question 34:

Differentiate

esin-1 2x

Answer:

Let  y=esin-12x
Differentiate it with respect to x we get,
dydx=ddxesin-12x       =esin-12x×ddxsin-12x             Using chain rule       =esin-12x×11-2x2ddx2x       =2esin-12x1-4x2So, ddxesin-12x=2esin-12x1-4x2

Page No 10.37:

Question 35:

Differentiate

sin 2 sin-1 x

Answer:

Let y=sin2 sin-1x
Differentiate it with respect to x we get,
dydx=ddxsin2sin-1x       =cos2 sin-1xddx2 sin-1x            Using chain rule       =cos2 sin-1x×211-x2       =2cos2 sin-1x1-x2So,ddxsin2sin-1x=2cos2 sin-1x1-x2

Page No 10.37:

Question 36:

Differentiate

etan-1 x

Answer:

Let y=etan-1x
Differentiate it with respect to x we get,
dydx=ddxetan-1x       =etan-1xddxtan-1x             Using chain rule       =etan-1x×11+x2ddxx       =etan-1x1+x×12x       =etan-1x2x1+xSo,  ddxetan-1x=etan-1x2x1+x

Page No 10.37:

Question 37:

Differentiate

tan-1x2

Answer:

Let  y=tan-1x2y=tan-1x212
Differentiate it with respect to x we get,
dydx=ddxtan-1x212       =12tan-1x212-1ddxtan-1x2                 Using chain rule       =12tan-1x2-12×11+x22×ddxx2       =444+x2tan-1x2       =14+x2tan-1x2So, ddxtan-1x2=14+x2tan-1<