Differentiate the function a 0 x n + a 1 x n - 1 + a 2 x n - 2 + … + a n - 1 x + a n w.r.t x .

Given, a 0 x n + a 1 x n - 1 + a 2 x n - 2 + … + a n - 1 x + a n Now, d d x a 0 x n + d d x a 1 x n - 1 + d d x a 2 x n - 2 + … + a n - 1 x + a n = a 0 d d x x n + a 1 d d x x n - 1 + a 2 d d x x n - 2 + … + a n - 1 d d x ( x ) + d d x a n = n a 0 x n - 1 + ( n - 1 ) a 1 x n - 2 + ( n - 2 ) a 2 x n - 3 + … . + a n - 1 ( 1 ) + 0 = n a 0 x n - 1 + ( n - 1 ) a 1 x n - 2 + ( n - 2 ) a 2 x n - 3 + … + a n - 1

Differentiate the function 1 sin x + 2 x + 3 + 4 log x 3 w.r.t x .

Given, 1 sin x + 2 x + 3 + 4 log x 3 Now, d d x 1 sin x + 2 x + 3 + 4 log x 3 = d d x cosec x + 2 x · 2 3 + 4 log 3 3 log x = d d x ( cosec x ) + 2 3 d d x 2 x + 4 log 3 d d x ( log x ) = - cosec x cot x + 2 3 · 2 x · log 2 + 4 log 3 · 1 x = - cosec x cot x + 2 x + 3 · log 2 + 4 x log 3

Differentiate the function ( x + 5 ) 2 x 2 - 1 x w.r.t x .

Given, ( x + 5 ) 2 x 2 - 1 x Now, d d x ( x + 5 ) 2 x 2 - 1 x = d d x 2 x 3 + 10 x 2 - x - 5 x = d d x 2 x 3 x + d d x 10 x 2 x - d d x x x - d d x 5 x = 2 d d x x 2 + 10 d d x ( x ) - d d x ( 1 ) - 5 d d x x - 1 = 2 ( 2 x ) + 10 ( 1 ) - 0 - 5 ( - 1 ) x - 2 = 4 x + 10 + 5 x 2

Differentiate the function log 1 x + 5 x a - 3 a x + x 2 3 + 6 x - 3 4 w.r.t x .

Given, log 1 x + 5 x a - 3 a x + x 2 3 + 6 x - 3 4 Now, d d x log 1 x + 5 x a - 3 a x + x 2 3 + 6 x - 3 4 = d d x log x - 1 2 + 5 d d x x a - 3 d d x a x + d d x x 2 3 + 6 d d x x - 3 4 = d d x - 1 2 log x + 5 d d x x a - 3 d d x a x + d d x x 2 3 + 6 d d x x - 3 4 = - 1 2 · 1 x + 5 a x a - 1 - 3 a x log a + 2 3 x - 1 3 + 6 - 3 4 x - 7 4 = - 1 2 x + 5 a x a - 1 - 3 a x log a + 2 3 x - 1 3 - 9 2 x - 7 4

Differentiate the function cos ( x - 2 ) sin x w.r.t x

Given, cos ( x - 2 ) sin x Now, d d x cos ( x - 2 ) sin x = d d x cos x cos 2 + sin x sin 2 sin x = d d x cos x cos 2 sin x + d d x sin x sin 2 sin x = cos 2 d d x ( cot x ) + sin 2 d d x ( 1 ) = cos 2 - cosec 2 x + sin 2 ( 0 ) = - cosec 2 x cos 2

If y = sin x 2 + cos x 2 2 , find d y d x at x = π 6

We have to find derivative of d y d x = d d x sin x 2 + cos x 2 2 = d d x sin 2 x 2 + cos 2 x 2 + 2 sin x 2 cos x 2 = d d x ( 1 + sin x ) = d d x ( 1 ) + d d x ( sin x ) = 0 + cos x ∴ d y d x = cos x d y d x at x = π 6 is = cos π 6 = 3 2

If y = 2 - 3 cos x sin x , find d y d x at x = π 4

d y d x = d d x 2 - 3 cos x sin x = d d x 2 sin x - d d x 3 cos x sin x = 2 d d x ( cosec x ) - 3 d d x ( cot x ) = - 2 cosec x cot x + 3 cosec 2 x d y d x at x = π 4 is = - 2 cosec π 4 cot π 4 + 3 cosec 2 π 4 = - 2 ( 2 ) ( 1 ) + 3 ( 2 ) 2 = - 2 2 + 6 = 6 - 2 2

E M B I B E

MATHEMATICS for CLASS XI>Chapter 30 - Derivatives>EXERCISE 30.3

R D Sharma Solutions for Chapter: Derivatives, Exercise 3: EXERCISE 30.3

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Derivatives, Exercise 3: EXERCISE 30.3

Attempt the practice questions on Chapter 30: Derivatives, Exercise 3: EXERCISE 30.3 with hints and solutions to strengthen your understanding. MATHEMATICS for CLASS XI solutions are prepared by Experienced Embibe Experts.