Differentiate tan - 1 x - 1 x + 1 with respect to sin - 1 3 x - 4 x 3 if - 1 2 < x < 1 2

Let u = tan - 1 x - 1 x + 1 Put x = tan θ So we will get, , u = tan - 1 tan θ - 1 tan θ + 1 ⇒ u = tan - 1 tan θ - tan π 4 1 + tan θ tan π 4 ⇒ u = tan - 1 tan θ - π 4 . . . . . . . i Here - 1 2 < x < 1 2 ⇒ - 1 2 < tan θ < 1 2 ⇒ - tan - 1 1 2 < θ < tan - 1 1 2 So from equation i u = θ - π 4 ∵ tan - 1 tan θ = θ , θ ∈ - π 2 , π 2 u = tan - 1 x - π 4 Differentiating both side w.r.t x we get d u d x = d d x tan - 1 x - π 4 ⇒ d u d x = 1 1 + x 2 . . . . . . . . ii Similarly let, v = sin - 1 3 x - 4 x 3 put x = sin θ v = sin - 1 3 sin θ - 4 sin θ 3 ⇒ v = sin - 1 sin 3 θ . . . . . . . . . iii Here, - 1 2 < x < 1 2 ⇒ - 1 2 < sin θ < 1 2 ⇒ - π 6 < θ < π 6 From equation iii v = 3 θ ∵ sin - 1 sin θ = θ , θ ∈ - π 2 , π 2 v = 3 sin - 1 x Differentiating both side w.r.t x we get d v d x = d d x 3 sin - 1 x ⇒ d v d x = 3 1 - x 2 . . . . . . . . iv Dividing equation iii by iv d u d x d v d x = 1 1 + x 2 × 1 - x 2 3 ⇒ d u d x = 1 - x 2 3 1 + x 2

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

Let u = sin - 1 2 x 1 + x 2 put x = tan θ So, u = sin - 1 2 x 1 + x 2 ⇒ u = sin - 1 2 tan θ 1 + tan 2 θ ⇒ u = sin - 1 sin 2 θ . . . . . . . . i Let, v = tan - 1 2 x 1 - x 2 Put x = tan θ v = tan - 1 2 x 1 - x 2 ⇒ v = tan - 1 2 tan θ 1 - tan 2 θ ⇒ v = tan - 1 tan 2 θ . . . . . . . . . ii Here - 1 < x < 1 ⇒ - 1 < sin θ < 1 ⇒ - π 2 < θ < π 2 So from equation i we get u = 2 θ ∵ sin - 1 sin θ = θ , θ ∈ - π 2 , π 2 u = 2 tan - 1 x . . . . . iii Similarly here - 1 < tan θ < 1 ⇒ - π 4 < θ < π 4 So from equation ii we get v = 2 tan - 1 x . . . . . iv Differentiating equation iii and iv w.r.t x d u d x = d d x 2 tan - 1 x ⇒ d u d x = 2 1 + x 2 . . . . . . . . v Similarly, d v d x = d d x 2 tan - 1 x ⇒ d v d x = 2 1 + x 2 . . . . . . . vi From equation v and vi d u d x d v d x = 2 1 + x 2 × 1 + x 2 2 ⇒ d u d v = 1

Differentiate cos - 1 4 x 3 - 3 x with respect to tan - 1 1 - x 2 x if 1 2 < x < 1

Let u = cos - 1 4 x 3 - 3 x put x = cos θ So, u = cos - 1 4 x 3 - 3 x ⇒ u = cos - 1 4 cos 3 θ - 3 cos θ ⇒ u = cos - 1 cos 3 θ . . . . . . . i Here 1 2 < x < 1 ⇒ 1 2 < cos θ < 1 ⇒ 0 < θ < π 3 So, from equation i u = 3 θ ∵ cos - 1 cos θ = θ , θ ∈ 0 , π u = 3 cos - 1 x Differentiating both side w.r.t x we get d u d x = d d x 3 cos - 1 x ⇒ d u d x = - 3 1 - x 2 . . . . . . . ii Similarly let v = tan - 1 1 - x 2 x Put x = cos θ So, v = tan - 1 1 - cos 2 θ cos θ ⇒ v = tan - 1 sin θ cos θ ⇒ v = tan - 1 tan θ . . . . . . . . . iii Here 1 2 < x < 1 ⇒ 1 2 < tan θ < 1 So from equation iii v = θ ∵ tan - 1 tan θ = θ , θ ∈ - π 2 , π 2 v = cos - 1 x Differentiating both side w.r.t x we get d v d x = d d x cos - 1 x ⇒ d v d x = - 1 1 - x 2 . . . . . . . . . iv From equation iii and iv d u d x d v d x = - 3 1 - x 2 - 1 - x 2 1 ⇒ d u d v = 3

Differentiate tan - 1 x 1 - x 2 with respect to sin - 1 2 x 1 - x 2 , if - 1 2 < x < 1 2 ?

Let, u = tan - 1 x 1 - x 2 Put x = sin θ ⇒ θ = sin - 1 x ⇒ u = tan - 1 sin θ 1 - sin 2 θ ⇒ u = tan - 1 sin θ cos θ ⇒ u = tan - 1 tan θ .......... i And Let, v = sin - 1 2 x 1 - x 2 v = sin - 1 2 sin θ 1 - sin 2 θ v = sin - 1 2 sin θ cos θ v = sin - 1 sin 2 θ ......... ii Here, - 1 2 < x < 1 2 ⇒ - 1 2 < sin θ < 1 2 ⇒ - π 4 < θ < π 4 So, from equation i , u = θ ∵ tan - 1 tan θ = θ , θ ∈ - π 2 , π 2 ⇒ u = sin - 1 x Differentiating it with respect to x , d u d x = 1 1 - x 2 ........ iii from equation ii , v = 2 θ ∵ sin - 1 sin θ = θ , θ ∈ - π 2 , π 2 ⇒ v = 2 sin - 1 x Differentiating it with respect to x , d v d x = 2 1 - x 2 ........ iv Dividing equation iii by iv d u d x d v d x = 1 1 - x 2 1 - x 2 2 ∴ d u d v = 1 2

sin - 1 1 - x 2 with respect to cot - 1 x 1 - x 2 , if 0 < x < 1 ?

Let, u = sin - 1 1 - x 2 Put x = cos θ ⇒ θ = cos - 1 x We get, u = sin - 1 1 - cos 2 θ ⇒ u = sin - 1 sin θ ...... i Let, v = cot - 1 x 1 - x 2 ⇒ v = cot - 1 cos θ 1 - cos 2 θ ⇒ v = cot - 1 cos θ sin θ ⇒ v = cot - 1 cot θ ...... ii Here, 0 < x < 1 ⇒ 0 < cos θ < 1 ⇒ 0 < θ < π 2 So, from equation i , u = θ ∵ sin - 1 sin θ = θ , θ ∈ - π 2 , π 2 ⇒ u = cos - 1 x Differentiating it with respect to x , d u d x = - 1 1 - x 2 .... iii From equation ii v = θ ∵ cot - 1 cot θ = θ , θ ∈ 0 , π ⇒ v = cos - 1 x Differentiating it with respect to x , d v d x = - 1 1 - x 2 ..... iv Dividing equation (iii) by (iv), d u d x d v d x = - 1 1 - x 2 1 - x 2 - 1 ∴ d u d v = 1

Differentiate sin - 1 2 a x 1 - a 2 x 2 with respect to 1 - a 2 x 2 if - 1 2 < a x < 1 2

Let, u = sin - 1 2 a x 1 - a 2 x 2 Put a x = sin θ ∴ θ = sin - 1 a x u = sin - 1 2 sin θ 1 - sin 2 θ ⇒ u = sin - 1 2 sin θ cos θ ⇒ u = sin - 1 sin 2 θ Here given, - 1 2 < a x < 1 2 - 1 2 < sin θ < 1 2 - π 4 < θ < π 4 So, u = 2 θ ∵ sin - 1 sin θ = θ , θ ∈ - π 2 , π 2 u = 2 sin - 1 x Differentiating both side w.r.t x we get d u d x = d d x 2 sin - 1 x ⇒ d u d x = 2 1 - a x 2 d d x a x ⇒ d u d x = 2 a 1 - a x 2 Let v = 1 - a 2 x 2 Differentiating both side w.r.t x we get d v d x = d d x 1 - a 2 x 2 ⇒ d v d x = 1 2 1 - a 2 x 2 d d x 1 - a 2 x 2 ⇒ d v d x = - 2 a 2 x 2 1 - a 2 x 2 ⇒ d v d x = - a 2 x 1 - a 2 x 2 So d u d v = d u d x × d x d v d u d v = 2 a 1 - a x 2 - 1 - a 2 x 2 a 2 x ⇒ d u d v = - 2 a x

Differentiate tan - 1 1 - x 1 + x with respect to 1 - x 2 , if - 1 < x < 1 ?

Let, u = tan - 1 1 - x 1 + x Put x = tan θ ⇒ θ = tan - 1 x ⇒ u = tan - 1 1 - tan θ 1 + tan θ ⇒ u = tan - 1 tan π 4 - θ ......... i Here, - 1 < x < 1 ⇒ - 1 < tan θ < 1 ⇒ - π 4 < θ < π 4 ⇒ π 4 > - θ > π 4 ⇒ - π 4 < - θ < π 4 ⇒ 0 < π 4 - θ < π 2 So, from equation i u = π 4 - θ ∵ tan - 1 tan θ = θ , θ ∈ - π 2 , π 2 ⇒ u = π 4 - tan - 1 x Differentiating it with respect to x , d u d x = 0 - 1 1 + x 2 ⇒ d u d x = - 1 1 + x 2 ...... ii let, v = 1 - x 2 Differentiating it with respect to x , d v d x = 1 2 1 - x 2 × d d x 1 - x 2 ⇒ d v d x = 1 2 1 - x 2 - 2 x ⇒ d v d x = - x 1 - x 2 .......... iii Dividing equation ii by iii , d u d x d v d x = - 1 1 + x 2 × 1 - x 2 - x ∴ d u d v = 1 - x 2 x 1 + x 2

E M B I B E

MATHEMATICS CLASS XII VOLUME-1>Chapter 11 - Differentiation>EXERCISE

R D Sharma Solutions for Chapter: Differentiation, Exercise 1: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Differentiation, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 11: Differentiation, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-1 solutions are prepared by Experienced Embibe Experts.

Questions from R D Sharma Solutions for Chapter: Differentiation, Exercise 1: EXERCISE with Hints & Solutions

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 11 - Differentiation>EXERCISE>Q 1

Differentiate $x^{2}$ with respect to $x^{3}$ .

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 11 - Differentiation>EXERCISE>Q 13

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

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 11 - Differentiation>EXERCISE>Q 15

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

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 11 - Differentiation>EXERCISE>Q 16

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

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 11 - Differentiation>EXERCISE>Q 17

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

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 11 - Differentiation>EXERCISE>Q 18

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

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 11 - Differentiation>EXERCISE>Q 19

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