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

Let, y = tan - 1 x a + a 2 - x 2 Put x = a sin θ ⇒ y = tan - 1 a sin θ a + a 2 - a 2 sin 2 θ ⇒ y = tan - 1 a sin θ a + a 2 1 - sin 2 θ ⇒ y = tan - 1 a sin θ a + a cos θ ⇒ y = tan - 1 a sin θ a ( 1 + cos θ ) ⇒ y = tan - 1 sin θ 1 + cos θ We know 1 + cos θ = 2 cos 2 θ 2 and sin x = 2 sin θ 2 cos θ 2 So, y = tan - 1 2 sin θ 2 cos θ 2 2 cos 2 θ 2 ⇒ y = tan - 1 sin θ 2 cos θ 2 ⇒ y = tan - 1 tan θ 2 Here - a < x < a - 1 < x a < 1 - 1 < sin θ < 1 - π 2 < θ < π 2 - π 4 < θ 2 < π 4 So y = θ 2 ∵ tan - 1 tan θ = θ , θ ∈ - π 2 , π 2 y = 1 2 sin - 1 x a Differentiating both side w.r.t x we get d y d x = d d x 1 2 sin - 1 x a ⇒ d y d x = 1 2 1 1 - x a 2 d d x x a ⇒ d y d x = 1 2 a a 2 - x 2 1 a ⇒ d y d x = 1 2 a 2 - x 2

Differentiate sin - 1 x + 1 - x 2 2 , - 1 < x < 1 with respect to x .

Let, y = sin - 1 x + 1 - x 2 2 Put x = sin θ ∴ y = sin - 1 sin θ + 1 - sin 2 θ 2 ⇒ y = sin - 1 sin θ + cos θ 2 ⇒ y = sin - 1 sin θ 1 2 + cos θ 1 2 ⇒ y = sin - 1 sin θ cos π 4 + cos θ sin π 4 ⇒ y = sin - 1 sin θ + π 4 Here - 1 < x < 1 - 1 < sin θ < 1 - π 2 < θ < π 2 - π 2 + π 4 < θ + π 4 < π 2 + π 4 - π 4 < θ + π 4 < 3 π 4 y = θ + π 4 ∵ sin - 1 sin θ = θ , θ ∈ - π 2 , π 2 y = sin - 1 x + π 4 Differentiating both side w.r.t x we get d y d x = d d x sin - 1 x + π 4 ⇒ d y d x = 1 1 - x 2

Differentiate cos - 1 x + 1 - x 2 2 , - 1 < x < 1 with respect to x

Let y = cos - 1 x + 1 - x 2 2 Put , x = cos θ y = cos - 1 cos θ + 1 - cos 2 θ 2 y = cos - 1 cos θ + sin θ 2 y = cos - 1 cos θ 1 2 + sin θ 1 2 y = cos - 1 cos θ cos π 4 + sin θ sin π 4 y = cos θ cos π 4 + sin θ sin π 4 y = cos θ - π 4 Here - 1 < x < 1 - 1 < cos θ < 1 3 π 4 < θ < 5 π 4 3 π 4 - π 4 < θ - π 4 < 5 π 4 - π 4 π 2 < θ + π 4 < π So y = θ - π 4 ∵ cos - 1 cos θ = θ , θ ∈ 0 , π y = cos - 1 x - π 4 Differentiating both side w.r.t x we get d y d x = d d x cos - 1 x - π 4 ⇒ d y d x = - 1 1 - x 2

Differentiate tan - 1 4 x 1 - 4 x 2 , - 1 2 < x < 1 2 with respect to x .

Let, y = tan - 1 4 x 1 - 4 x 2 put 2 x = tan θ ⇒ y = tan - 1 2 tan θ 1 - tan 2 θ ⇒ y = tan - 1 ( tan 2 θ ) Here, - 1 2 < x < 1 2 ⇒ - 1 < 2 x < 1 ⇒ - 1 < tan θ < 1 ⇒ - π 4 < θ < π 4 ⇒ - π 2 < 2 θ < π 2 So, y = 2 θ ∵ tan - 1 tan θ = θ , θ ∈ - π 2 , π 2 y = 2 tan - 1 2 x Differentiating both side w.r.t x we get d y d x = d d x 2 tan - 1 2 x ⇒ d y d x = 2 1 + 2 x 2 d d x 2 x ⇒ d y d x = 4 1 + 4 x 2

If y = cos - 1 2 x - 3 1 - x 2 13 , find d y d x ?

Let. y = cos - 1 2 x - 3 1 - x 2 13 Put, x = cos θ y = cos - 1 2 cos θ - 3 1 - cos 2 θ 13 y = cos - 1 2 cos θ - 3 sin θ 13 y = cos - 1 cos θ 2 13 + sin θ - 3 13 Let, cos ϕ = 2 13 and sin ϕ = - 3 13 y = cos - 1 cos θ cos ϕ + sin θ sin ϕ y = cos - 1 cos θ - ϕ . . . i y = θ - ϕ y = cos - 1 x - ϕ Differentiating it with respect to x , d y d x = - 1 1 - x 2 + 0 d y d x = - 1 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

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 11 - Differentiation>EXERCISE>Q 12

If $y = \tan^{- 1} (\frac{1 + \cos x}{\sin x})$ then find the value of $\frac{d y}{d x}$

MEDIUM

12th CBSE

IMPORTANT

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

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

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 11 - Differentiation>EXERCISE>Q 14

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

MEDIUM

12th CBSE

IMPORTANT

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

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

MEDIUM

12th CBSE

IMPORTANT

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

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

MEDIUM

12th CBSE

IMPORTANT

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

If differentiation of $\tan^{- 1} (\frac{2^{x + 1}}{1 - 4^{x}}), - \infty < x < 0$ , with respect to $x$ is $\frac{a^{x + 1} \ln (b)}{c + d^{x}}$ , then $a + b + c + d =$