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)

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

Author:Telangana Board

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

Attempt the free 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 Telangana Board Solutions for Chapter: Differentiation, Exercise 3: Exercise 9(c) with Hints & Solutions

EASY

11th Telangana 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 Telangana 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 Telangana 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 Telangana 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 Telangana 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 $f^{'} (x) = (a + b \cos x)^{- 1}$ .