Write the value of lim x → c f ( x ) - f ( c ) x - c .

Using the definition of derivative, we have: lim x → c f ( x ) - f ( c ) x - c = f ' c Hence, the answer is f ' ( c ) .

Write the value of d d x x | x | .

Given, d d x x x Case (i): x > 0 , | x | = x ⇒ d d x ( x | x | ) = d d x ( x . x ) = d d x x 2 = 2 x Case (ii): x < 0 , | x | = - x ⇒ d d x x | x | = d d x x - x = d d x - x 2 = - 2 x From case (i) and case(ii), we have: d d x x | x | = 2 x , if x > 0 - 2 x , if x < 0

Write the value of d d x x + x x .

Given, d d x x + x x Case (i): x > 0 , | x | = x Now, ( x + | x | ) | x | = ( x + x ) x = 2 x 2 ⇒ d d x ( x + | x | ) | x | = d d x 2 x 2 = 4 x Case (ii): x < 0 , | x | = - x Now, ( x + | x | ) | x | = ( x - x ) x = 0 ⇒ d d x ( x + | x | ) | x | = d d x ( 0 ) = 0 From case (i) and case (ii), we have: d d x x + x x = 4 x , if x > 0 0 , if x < 0

If f ( x ) = | x | + | x - 1 | , write the value of d d x ( f ( x ) ) .

Given, f ( x ) = | x | + | x - 1 | Case (i): x < 0 ( ∴ x - 1 < - 1 < 0 ) | x | = - x ; | x - 1 | = - ( x - 1 ) = - x + 1 f ( x ) = - x + ( - x + 1 ) = - 2 x f ' ( x ) = - 2 Case (ii): 0 < x < 1 ( ∴ x > 0 and x - 1 < 0 ) | x | = x ; | x - 1 | = - ( x - 1 ) = 1 - x f ( x ) = x + 1 - x = 1 f ' ( x ) = 0 Case (iii): x > 1 ∴ x - 1 > 0 and x > 0 | x | = x ; | x - 1 | = x - 1 f ( x ) = x + x - 1 = 2 x - 1 f ' ( x ) = 2 From case (i), case (ii) and case (iii), we have: f ' x = - 2 , when x < 0 0 , when 0 < x < 1 2 , when x > 1

If f x = x 2 | x | , write d d x f x .

Given, f x = x 2 x Case (i): x > 0 , | x | = x f x = x 2 | x | = x 2 x = x f ' ( x ) = 1 Case (ii): x < 0 , | x | = - x f ( x ) = x 2 | x | = x 2 - x = - x f ' ( x ) = - 1 From case (i) and case (ii), we have f ' x = 1 , if x > 0 - 1 , if x < 0

Write the value of d d x ( log | x | )

Case (i): x > 0 , x = x ...(i) d d x log | x | = d d x log x = 1 x = 1 | x | (from (i)) Case (ii): x < 0 , x = - x ...(ii) d d x log | x | = d d x log - x = 1 - x = 1 | x | (from (ii)) From case (i) and case (ii) d d x log | x | = 1 | x | , x ≠ 0 .

If | x | < 1 and y = 1 + x + x 2 + x 3 + … , then write the value of d y d x .

The given series is a geometric series where a = 1 and r = x and also x < 1 . f x = 1 + x + x 2 + x 3 + … = 1 1 - x Sum of the infinite terms of a geometric progression is a 1 - r . f ' x = - 1 ( 1 - x ) 2 d d x 1 - x = - 1 ( 1 - x ) 2 - 1 = 1 ( 1 - x ) 2

E M B I B E

MATHEMATICS for CLASS XI>Chapter 30 - Derivatives>VERY SHORT ANSWER QUESTIONS (VSAQs)

R D Sharma Solutions for Chapter: Derivatives, Exercise 6: VERY SHORT ANSWER QUESTIONS (VSAQs)

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Derivatives, Exercise 6: VERY SHORT ANSWER QUESTIONS (VSAQs)

Attempt the free practice questions on Chapter 30: Derivatives, Exercise 6: VERY SHORT ANSWER QUESTIONS (VSAQs) with hints and solutions to strengthen your understanding. MATHEMATICS for CLASS XI solutions are prepared by Experienced Embibe Experts.