If x y + y x = a b , then prove that d y d x = - y x y - 1 + y x log y x y log x + x y x - 1 .

Given, x y + y x = a b Let, u = x y and v = y x so that u + v = a b Now, u = x y Applying log , ⇒ log u = log x y ⇒ log u = y log x Differentiating w.r.t x , we get 1 u · d u d x = y · 1 x + log x · d y d x , ⇒ d u d x = u y x + log x · d y d x = x y y x + log x · d y d x Now, v = y x Applying log , ⇒ log v = log y x ⇒ log v = x log y ⇒ 1 v · d v d x = x · 1 y · d y d x + log y . 1 ⇒ d v d x = v x y · d y d x + log y = y x x y · d y d x + log y Since, u + v = a b ⇒ d u d x + d v d x = 0 ⇒ x y y x + log x d y d x + y x x y · d y d x + log y = 0 ⇒ y x y - 1 + x y · log x d y d x + x · y x - 1 d y d x + y x · log y = 0 ⇒ d y d x · x y · log x + x · y x - 1 = - y · x y - 1 + y x · log y ∴ d y d x = - y · x y - 1 + y x · log y x y · log x + x · y x - 1

If f x = sin - 1 x - β α - β and g x = tan - 1 x - β α - x , then prove that f ' ( x ) = g ' ( x ) ( β < x < α ) .

Given, f x = sin - 1 x - β α - β Let y = sin - 1 x - β α - β ⇒ sin y = x - β α - β ⇒ sin 2 y = x - β α - β We know that, cos 2 y = 1 - sin 2 y ⇒ cos 2 y = 1 - x - β α - β = α - β - x + β α - β = α - x α - β ⇒ cos y = α - x α - β ∴ tan y = sin y cos y = x - β α - β α - x α - β = x - β α - β · α - β α - x = x - β α - x y = tan - 1 x - β α - x ⇒ sin - 1 x - β α - x = tan - 1 x - β α - x ∴ f ' ( x ) = g ' ( x ) Hence, proved.

If a > b > 0 and 0 < x < π ; f x = a 2 - b 2 - 1 2 · cos - 1 a cos x + b a + b cos x , then prove that f ' ( x ) = ( a + b cos x ) - 1 .

Let, u = cos - 1 a cos x + b a + b cos x ⇒ d u d x = - 1 1 - ( a cos x + b ) 2 ( a + b cos x ) 2 · d d x a cos x + b a + b cos x = - 1 1 - ( a cos x + b ) 2 ( a + b cos x ) 2 · ( a + b cos x ) ( - a sin x ) - ( a cos x + b ) ( - b sin x ) ( a + b cos x ) 2 = - a + b cos x ( a + b cos x ) 2 - ( a cos x + b ) 2 × ( a + b cos x ) ( - a sin x ) - ( a cos x + b ) ( - b sin x ) ( a + b cos x ) 2 = a 2 - b 2 sin x ( a + b cos x ) a 2 - b 2 1 - cos 2 x = a 2 - b 2 sin x ( a + b cos x ) a 2 - b 2 · sin x = a 2 - b 2 a + b cos x f ' x = a 2 - b 2 - 1 2 a 2 - b 2 a + b cos x = 1 a + b cos x ∴ f ' x = ( a + b cos x ) - 1 Hence, proved.

E M B I B E

Intermediate First Year Mathematics Paper 1B>Chapter 9 - Differentiation>Exercise 9(c)

Andhra Pradesh Board Solutions for Chapter: Differentiation, Exercise 3: Exercise 9(c)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Chapter: Differentiation, Exercise 3: Exercise 9(c)

Attempt the practice questions on Chapter 9: Differentiation, Exercise 3: Exercise 9(c) with hints and solutions to strengthen your understanding. Intermediate First Year Mathematics Paper 1B solutions are prepared by Experienced Embibe Experts.

Questions from Andhra Pradesh Board Solutions for Chapter: Differentiation, Exercise 3: Exercise 9(c) with Hints & Solutions

EASY

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 9 - Differentiation>Exercise 9(c)>Q I

Find the derivative of the function $\tan^{- 1} \sqrt{\frac{1 - \cos x}{1 + \cos x}}$ .

EASY

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 9 - Differentiation>Exercise 9(c)>Q I

Differentiate $f (x)$ with respect to $g (x)$ where $f (x) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}), g (x) = \sin^{- 1} (\frac{2 x}{1 + x^{2}})$ .

HARD

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 9 - Differentiation>Exercise 9(c)>Q III

If $x^{y} + y^{x} = a^{b}$ , then prove that $\frac{d y}{d x} = - [\frac{y x^{y - 1} + y^{x} \log y}{x^{y} \log x + x y^{x - 1}}]$ .

MEDIUM

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 9 - Differentiation>Exercise 9(c)>Q III

If $f (x) = \sin^{- 1} \sqrt{\frac{x - β}{α - β}}$ and $g (x) = \tan^{- 1} \sqrt{\frac{x - β}{α - x}}$ , then prove that $f^{'} (x) = g^{'} (x) (β < x < α)$ .

MEDIUM

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 9 - Differentiation>Exercise 9(c)>Q III

If $a > b > 0$ and $0 < x < π; f (x) = {(a^{2} - b^{2})}^{- \frac{1}{2}} \cdot \cos^{- 1} (\frac{a \cos x + b}{a + b \cos x})$ , then prove that $f^{'} (x) = (a + b \cos x)^{- 1}$ .