If x = a sin 2 t ( 1 + cos 2 t ) , y = b cos 2 t ( 1 - cos 2 t ) , show that d y d x = b a at t = π 4

Given x = a sin 2 t ( 1 + cos 2 t ) and y = b cos 2 t ( 1 - cos 2 t ) Now, d x d t = a sin 2 t · d d t ( 1 + cos 2 t ) + ( 1 + cos 2 t ) · d d t sin 2 t = a sin 2 t ( - sin 2 t ) · d d t 2 t + ( 1 + cos 2 t ) · cos 2 t · d d t 2 t = - 2 a sin 2 2 t + 2 a cos 2 t ( 1 + cos 2 t ) and d y d t = b cos 2 t d d t ( 1 - cos 2 t ) + ( 1 - cos 2 t ) · d d t cos 2 t = b cos 2 t · ( sin 2 t ) d d t 2 t + ( 1 - cos 2 t ) ( - sin 2 t ) · d d t 2 t = b [ 2 sin 2 t · cos 2 t + 2 ( 1 - cos 2 t ) ( - sin 2 t ) ] = 2 b [ sin 2 t . cos 2 t - ( 1 - cos 2 t ) sin 2 t ] ∴ d y d x = d y / d t d x / d t ⇒ d y d x = - 2 b [ - sin 2 t · cos 2 t + ( 1 - cos 2 t ) sin 2 t ] - 2 a sin 2 2 t - cos 2 t ( 1 + cos 2 t ) ⇒ d y d x a t t = π 4 = b - sin π 2 cos π 2 + 1 - cos π 2 sin π 2 a sin 2 π 2 - cos π 2 1 + cos π 2 = b a · ( 0 + 1 ) ( 1 - 0 ) ∵ sin π 2 = 1 and cos π 2 = 0 = b a Hence proved.

If f x = 3 + x 1 + x 2 + 3 x ; find f ' 0 .

The given function is, f x = 3 + x 1 + x 2 + 3 x Let f x = y y = 3 + x 1 + x 2 + 3 x Taking log of both sides ⇒ log y = log 3 + x 1 + x 2 + 3 x ⇒ log y = 2 + 3 x log 3 + x 1 + x ∵ log m n = n log m On differentiating both side with respect to x , we get d d x log y = d d x 2 + 3 x log 3 + x 1 + x 1 y · d y d x = 2 + 3 x d d x log 3 + x 1 + x + log 3 + x 1 + x d d x 2 + 3 x [ Using product rule ] 1 y · d y d x = 2 + 3 x · 1 3 + x 1 + x d d x 3 + x 1 + x + log 3 + x 1 + x d d x 2 + d d x 3 x [ Use chain rule and also ∵ d d x log x = 1 x 1 y · d y d x = 2 + 3 x · 1 + x 3 + x 1 + x d d x 3 + x - 3 + x d d x 1 + x 1 + x 2 + log 3 + x 1 + x 0 + 3 [ Using quotient rule ] 1 y · d y d x = 2 + 3 x · 1 + x 3 + x 1 + x d d x 3 + d d x x - 3 + x d d x 1 + d d x x 1 + x 2 + 3 log 3 + x 1 + x ∵ d d x f x + g x = d d x f x + d d x g x 1 y · d y d x = 2 + 3 x · 1 + x 3 + x 1 + x 0 + 1 - 3 + x 0 + 1 1 + x 2 + 3 log 3 + x 1 + x ∵ d d x x n = n x n - 1 1 y · d y d x = 2 + 3 x · 1 + x 3 + x 1 + x - 3 + x 1 + x 2 + 3 log 3 + x 1 + x 1 y · d y d x = 2 + 3 x · 1 + x 3 + x 1 + x - 3 - x 1 + x 2 + 3 log 3 + x 1 + x 1 y · d y d x = 2 + 3 x · 1 + x 3 + x - 2 1 + x 2 + 3 log 3 + x 1 + x d y d x = y · 2 + 3 x · 1 + x 3 + x - 2 1 + x 2 + 3 log 3 + x 1 + x [ Cross multiply y ] Put d y d x = f ' x f ' x = 3 + x 1 + x 2 + 3 x · 2 + 3 x · 1 + x 3 + x - 2 1 + x 2 + 3 log 3 + x 1 + x For f ' 0 f ' 0 = 3 + 0 1 + 0 2 + 3 × 0 · 2 + 3 × 0 · 1 + 0 3 + 0 - 2 1 + 0 2 + 3 log 3 + 0 1 + 0 f ' 0 = 3 2 · 2 · 1 3 · - 2 + 3 log 3 f ' 0 = 9 · - 4 3 + 3 log 3 f ' 0 = 9 × - 4 3 + 9 × 3 log 3 f ' 0 = - 12 + 27 log 3 .

Find d y d x , if 2 x + 3 y = sin y .

Given, 2 x + 3 y = sin y Differentiating both sides with respect to x , we get d d x 2 x + 3 y = d d x sin y ⇒ d d x 2 x + d d x 3 y = cos y d y d x [Using addition rule of differentiation] ⇒ 2 + 3 d y d x = cos y d y d x ∵ d d x x n = n x n - 1 ⇒ cos y d y d x - 3 d y d x = 2 ⇒ cos y - 3 d y d x = 2 ⇒ d y d x = 2 cos y - 3 .

If y = x sin x + ( sin x ) cos x , find d y d x .

Given, y = x sin x + ( sin x ) cos x Let u = x sin x , v = ( sin x ) cos x So, y = u + v ⇒ d y d x = d u d x + d v d x .....(1) Consider, u = x sin x Taking log both sides, we get log u = sin x log x ∵ log a b = b log a On differentiating both sides w.r.t. x , we get d d x log u = log x · d d x ( sin x ) + sin x · d d x log x ⇒ 1 u d u d x = log x · ( cos x ) + sin x · 1 x ⇒ d u d x = u log x ( cos x ) + sin x x = x sin x log x ( cos x ) + sin x x .....(2) Consider, v = ( sin x ) cos x Taking log both sides, we get log v = cos x log sin x ∵ log a b = b log a On differentiating both sides w.r.t. x , we get d d x log v = d d x ( cos x ) · log ( sin x ) + cos x · d d x log ( sin x ) ⇒ 1 v d v d x = ( - sin x ) · log ( sin x ) + cos x · 1 sin x d d x ( sin x ) ⇒ 1 v d v d x = ( - sin x ) · log ( sin x ) + cos x · 1 sin x · cos x ⇒ 1 v d v d x = - y sin 2 x log ( sin x ) + cos 2 x sin x ⇒ d v d x = - sin x cos x sin 2 x log ( sin x ) + cos 2 x sin x .....(3) From (1), (2) and (3), we get d y d x = x sin x log x ( cos x ) + sin x x - sin x cos x sin 2 x log ( sin x ) + cos 2 x sin x

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 5 - Continuity and Differentiability>Nagaland Board-2018

Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 1: Nagaland Board-2020

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 1: Nagaland Board-2020

Attempt the free practice questions on Chapter 5: Continuity and Differentiability, Exercise 1: Nagaland Board-2020 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 1: Nagaland Board-2020 with Hints & Solutions

MEDIUM

12th Nagaland Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 5 - Continuity and Differentiability>Nagaland Board-2019>Q 9

If $y = a \cos (\log x) + b \sin (\log x)$ , prove that $x^{2} y_{2} + x y_{1} + y = 0$

MEDIUM

12th Nagaland Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 5 - Continuity and Differentiability>Nagaland Board-2019>Q 10

If $x = a \sin 2 t (1 + \cos 2 t), y = b \cos 2 t (1 - \cos 2 t)$ , show that $\frac{d y}{d x} = \frac{b}{a}$ at $t = \frac{π}{4}$

HARD

12th Nagaland Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 5 - Continuity and Differentiability>Nagaland Board-2019>Q 11

If $f (x) = {(\frac{3 + x}{1 + x})}^{2 + 3 x}$ ; find $f' (0)$ .

EASY

12th Nagaland Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 5 - Continuity and Differentiability>Nagaland Board-2020>Q 12

If $y = 5 \cos x - 3 \sin x$ , then $\frac{d^{2} y}{d x^{2}} + y$ is equal to

MEDIUM

12th Nagaland Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 5 - Continuity and Differentiability>Nagaland Board-2020>Q 13

Determine the value of $k$ so that the given function $f (x) = \{\begin{cases} \frac{x^{2} - 9}{x - 3}, x \neq 3 \\ k, if x = 3 \end{cases}$ is continuous at $x = 3$