Differentiate : 1 - tan x 1 + tan x

We have to differentiate 1 - tan x 1 + tan x . So, d d x 1 - tan x 1 + tan x = 1 + tan x d d x 1 - tan x - 1 - tan x d d x 1 + tan x 1 + tan x 2 [Using Quotient rule of differentiation] = 1 + tan x - sec 2 x - 1 - tan x sec 2 x 1 + tan x 2 = - sec 2 x - tan x sec 2 x - sec 2 x + tan x sec 2 x 1 + tan x 2 ∴ d d x 1 - tan x 1 + tan x = - 2 sec 2 x 1 + tan x 2 .

Differentiate : x 3 5 - 2 e 2 ln x + ln x 2 3 1 + x

Let y = x 3 5 - 2 e 2 ln x + ln x 2 3 1 + x y = x 3 5 - 2 e ln x 2 + 2 3 ln x 1 + x y = x 3 5 - 2 x 2 + 2 3 ln x 1 + x ∵ e ln a = a So, d y d x = d d x x 3 5 - 2 x 2 + 2 3 ln x 1 + x = 1 + x d d x x 3 5 - 2 x 2 + 2 3 ln x - x 3 5 - 2 x 2 + 2 3 ln x d d x 1 + x 1 + x 2 [Using Quotient rule of differentiation] = 1 + x 3 5 x - 2 5 - 4 x + 2 3 × 1 x - x 3 5 - 2 x 2 + 2 3 ln x 1 1 + x 2 ∴ d y d x = 1 + x 3 5 x - 2 5 - 4 x + 2 3 x - x 3 5 - 2 x 2 + 2 3 ln x 1 + x 2 .

Differentiate: a x - b x x

We have to differentiate a x - b x x . So, d d x a x - b x x = x d d x a x - b x - a x - b x d d x x x 2 [Using Quotient rule of differentiation] = x a x log a - b x log b - a x - b x 1 x 2 ∴ d d x a x - b x x = x a x log a - b x log b - a x - b x x 2 .

Differentiate: e x + e - x x 2 + 1

We have to differentiate e x + e - x x 2 + 1 . So, d d x e x + e - x x 2 + 1 = x 2 + 1 d d x e x + e - x - e x + e - x d d x x 2 + 1 x 2 + 1 2 [Using Quotient rule of differentiation] = x 2 + 1 e x + e - x d d x - x - e x + e - x d d x x 2 + 1 x 2 + 1 2 [Using chain rule of differentiation] = x 2 + 1 e x + e - x - 1 - e x + e - x 2 x x 2 + 1 2 ∴ d d x e x + e - x x 2 + 1 = x 2 + 1 e x - e - x - e x + e - x 2 x x 2 + 1 2 .

Differentiate : ln x x 2

We have to differentiate ln x x 2 . So, d d x ln x x 2 = x 2 d d x ln x - ln x d d x x 2 x 4 [Using Quotient rule of differentiation] = x 2 1 x - ln x 2 x x 4 ∴ d d x ln x x 2 = x - 2 x ln x x 4 .

E M B I B E

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 14 - Limit and Differentiation>EXERCISE 14(f)

Odisha Board Solutions for Chapter: Limit and Differentiation, Exercise 6: EXERCISE 14(f)

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Limit and Differentiation, Exercise 6: EXERCISE 14(f)

Attempt the free practice questions on Chapter 14: Limit and Differentiation, Exercise 6: EXERCISE 14(f) with hints and solutions to strengthen your understanding. Bureau's Higher Secondary Elements of Mathematics Vol.1 solutions are prepared by Experienced Embibe Experts.