Find the values of a so that the function f x defined by f x = sin 2 a x x 2 , x ≠ 0 1 , x = 0 may be continuous at x = 0 .

The function f x will be continuous at x = 0 , iff lim x → 0 f x = f 0 ⇒ lim x → 0 sin 2 a x x 2 = 1 ∵ f 0 = 1 ⇒ a 2 lim x → 0 sin a x a x 2 = 1 ⇒ a 2 1 2 = 1 ∵ lim z → 0 sin z z = 1 ⇒ a = ± 1

Find the derivative (if exists): sec - 1 1 8 x 2 - 1 w.r.t. 1 - x 2 at x = 1 2

Let the given functions be u = sec - 1 1 8 x 2 - 1 and v = 1 - x 2 We have to find, d u d v = d u d x d v d x d v d x = d 1 - x 2 d x = - 2 x 2 1 - x 2 = - x 1 - x 2 At x = 1 2 , d v d x x = 1 2 = - 1 2 1 - 1 4 = - 1 2 3 4 = - 1 3 Consider, u = sec - 1 1 8 x 2 - 1 We know that, d sec - 1 x d x = 1 x x 2 - 1 Now, d u d x = d sec - 1 1 8 x 2 - 1 d x Apply chain rule, d f g x d x = f ' g x g ' x ⇒ d u d x = 1 1 8 x 2 - 1 1 8 x 2 - 1 2 - 1 d 1 8 x 2 - 1 d x ⇒ d u d x = 1 1 8 x 2 - 1 1 8 x 2 - 1 2 - 1 × - 1 8 x 2 - 1 2 d 8 x 2 - 1 d x ⇒ d u d x = 1 1 8 x 2 - 1 1 8 x 2 - 1 2 - 1 × - 16 x 8 x 2 - 1 2 At x = 1 2 , 8 x 2 - 1 = 1 ⇒ d u d x x = 1 2 = ∞ ⇒ d u d v x = 1 2 = ∞ Hence it is undefined. So differentiation of given function is not exist.

Find dy dx when x y = e x .

Given that, x y = e x Taking log on both side we get, ⇒ log ( x y ) = log e x ⇒ y log ( x ) = x log e ⇒ y log ( x ) = x ⇒ y = x log x Differentiate w.r.t. x , ⇒ d y d x = d x log x d x ⇒ d y d x = log x d x d x - x d log x d x log x 2 ⇒ d y d x = log x - x 1 x log x 2 ⇒ d y d x = log x - 1 log x 2

Find dy dx when x y = y x

Given that, x y = y x Taking log both sides, ⇒ y log x = x log y Differentiate w.r.t. x , We know that, d d x u · v = u d v d x + v d u d x , d d x log x = 1 x and d d x x n = n x n - 1 . ⇒ y · 1 x + log x d y d x = x · 1 y · d y d x + log y d x d x ⇒ y x + log x d y d x = x y · d y d x + log y ⇒ log x - x y d y d x = log y - y x ⇒ y log x - x y d y d x = x log y - y x ⇒ d y d x = y x log y - y x y log x - x

E M B I B E

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Calculus - A. Continuity and Differentiability>Chapter 8 - Differentiation [Part-II]>EXERCISE

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 7: EXERCISE

Author:Parthasarathi Mukhopadhyay & Manabendra Nath Mukherjee

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 7: EXERCISE

Attempt the practice questions from Exercise 7: EXERCISE with hints and solutions to strengthen your understanding. RUDIMENTS of MATHEMATICS For Class 12 of +2 Level solutions are prepared by Experienced Embibe Experts.

Questions from Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 7: EXERCISE with Hints & Solutions

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 8 - Differentiation [Part-II]>EXERCISE>Q 3

The function $f (x) = | x |$ is

HARD

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 8 - Differentiation [Part-II]>EXERCISE>Q 18

Let $f (x - y) = \frac{f (x)}{f (y)}$ , for all real $x, y$ and $f (0) \neq 0, f^{'} (0) = p, f^{'} (5) = q$ . If the value of $f^{'} (- 5)$ is $\frac{p^{k}}{q}$ , then find the value of $k$ .