Find d y d x , when y = x 5 x + 4 ( 2 x + 3 ) 2

Given, y = x 5 x + 4 ( 2 x + 3 ) 2 Applying log on both sides, we get log y = 5 log x + 1 2 log x + 4 - 2 log 2 x + 3 ∵ log m n = log m + log n , log m n = log m - log n Now differentiating both sides by x , we get 1 y × d y d x = 5 x + 1 2 ( x + 4 ) - 4 2 x + 3 ⇒ d y d x = 5 x + 1 2 x + 8 - 4 2 x + 3 × y ∴ d y d x = 5 x + 1 2 x + 8 - 4 2 x + 3 x 5 x + 4 ( 2 x + 3 ) 2

Find d y d x , when y = ( x + 1 ) 2 x - 1 ( x + 4 ) 3 e x

Given, y = ( x + 1 ) 2 x - 1 ( x + 4 ) 3 e x Applying log on both sides, we get log y = 2 log x + 1 + 1 2 log x - 1 - 3 log x + 4 - x ∵ log m n = log m + log n , log m n = log m - log n , log e e = 1 Now differentiating both sides by x , we get 1 y × d y d x = 2 x + 1 + 1 2 ( x - 1 ) - 3 x + 4 - 1 ⇒ d y d x = 2 x + 1 + 1 2 ( x - 1 ) - 3 x + 4 - 1 × y ∴ d y d x = ( x + 1 ) 2 x - 1 ( x + 4 ) 3 e x 2 x + 1 + 1 2 ( x - 1 ) - 3 x + 4 - 1

Find d y d x , when y = x ( 3 x + 5 ) 2 x + 1

Given, y = x ( 3 x + 5 ) 2 x + 1 Taking log on both sides, we get log y = 2 log 3 x + 5 + 1 2 log x - log x + 1 ∵ log m n = log m + log n , log m n = log m - log n Now differentiating both sides by x , we get 1 y × d y d x = 2 3 x + 5 × 3 + 1 2 x - 1 2 ( x + 1 ) ⇒ d y d x = 6 3 x + 5 + 1 2 x - 1 2 ( x + 1 ) × y ∴ d y d x = ( 3 x + 5 ) 2 x x + 1 1 2 x + 6 3 x + 5 - 1 2 ( x + 1 )

Find d y d x , when y = x 2 1 + x 1 + x 2 3 2

Given, y = x 2 1 + x 1 + x 2 3 2 Taking log both the sides, we get log y = 2 log x + 1 2 log x + 1 - 3 2 log x 2 + 1 ∵ log m n = log m + log n , log m n = log m - log n Now differentiating both sides by x , we get 1 y × d y d x = 2 x + 1 2 ( x + 1 ) - 3 2 x 2 + 1 × 2 x ⇒ d y d x = 2 x + 1 2 ( x + 1 ) - 6 x 2 x 2 + 1 × y ∴ d y d x = x 2 x + 1 1 + x 2 3 2 2 x + 1 2 ( x + 1 ) - 3 x x 2 + 1

Find d y d x , when y = ( x - 2 ) ( 2 x - 3 ) ( 3 x - 4 )

Given, y = ( x - 2 ) ( 2 x - 3 ) ( 3 x - 4 ) Taking log on both sides, we get log y = 1 2 log x - 2 + log 2 x - 3 + log 3 x - 4 log m n = log m + log n , log m n = n log m Now differentiating both sides by x , we get 1 y × d y d x = 1 2 1 x - 2 + 2 2 x - 3 + 3 3 x - 4 ⇒ d y d x = 1 2 1 x - 2 + 2 2 x - 3 + 3 3 x - 4 × y ∴ d y d x = 1 2 ( x - 2 ) ( 2 x - 3 ) ( 3 x - 4 ) 1 x - 2 + 2 2 x - 3 + 3 3 x - 4

Find d y d x , when y = sin 2 x sin 3 x sin 4 x

Given, y = sin 2 x sin 3 x sin 4 x Taking, log on both sides, we get log y = log sin 2 x + log sin 3 x + log sin 4 x ∵ log m n = log m + log n Now differentiating both sides by x , we get 1 y × d y d x = 2 sin 2 x cos 2 x + 3 sin 3 x cos 3 x + 4 cot 4 x ⇒ d y d x = 2 cot 2 x + 3 cot 3 x + 4 cot 4 x × y ∴ d y d x = sin 2 x sin 3 x sin 4 x 2 cot 2 x + 3 cot 3 x + 4 cot 4 x

Find d y d x , when y = x 3 sin x e x

Given, y = x 3 sin x e x Applying log on both sides, we get log y = 3 log x + log sin x - x log m n = log m + log n , log m n = log m - log n , log e e = 1 Now differentiating both sides by x , we get 1 y × d y d x = 1 sin x × cos x + 3 x - 1 ⇒ d y d x = cos x sin x + 3 x - 1 × y ∴ d y d x = x 3 sin x e x 3 x + cot x - 1

Find d y d x , when y = e x log x x 2 .

Given, y = e x log x x 2 Now differentiating both sides by x , we get: d y d x = d d x e x log x x 2 ⇒ d y d x = x 2 d d x e x log x - e x log x d d x x 2 x 2 2 [Using Quotient Rule in Differentiation] ⇒ d y d x = x 2 e x d d x log x + log x d d x e x - e x log x d d x x 2 x 4 [Using Product Rule in Differentiation] ⇒ d y d x = x 2 e x × 1 x + log x e x - 2 x e x log x x 4 ⇒ d y d x = x e x + x 2 log x e x - 2 x e x log x x 4 ⇒ d y d x = x e x 1 + x log x - 2 log x x 4 ∴ d y d x = x e x 1 + x log x - 2 log x x 4 .

E M B I B E

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Differential Calculus>Chapter 10 - Differentiation>EXERCISE

R S Aggarwal Solutions for Exercise 6: EXERCISE

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 6: EXERCISE

Attempt the free practice questions from Exercise 6: EXERCISE with hints and solutions to strengthen your understanding. SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12 solutions are prepared by Experienced Embibe Experts.