If x = 4 z 2 + 5 , y = 6 z 2 + 7 z + 3 , find d 2 y d x 2

Here, x = 4 z 2 + 5 and y = 6 z 2 + 7 z + 3 Differentiating w.r.t. z, we get d x d z = 8 z and d y d z = 12 z + 7 ∴ d y d x = d y d z d x d z = 12 z + 7 8 z Differentiating w.r.t. x , we get d 2 y d x 2 = 12 × 8 z - 8 ( 12 z + 7 ) 8 z 2 × d z d x ⇒ d 2 y d x 2 = 12 × 8 z - 8 ( 12 z + 7 ) 64 z 2 × 1 8 z ⇒ d 2 y d x 2 = 96 z - 96 z - 56 512 z 3 = - 7 64 z 3

If x = cos t + log tan t 2 , y = sin t , then find the value of d 2 y d t 2 and d 2 y d x 2 at t = π 4 .

Given, x = cos t + log tan t 2 , y = sin t ⇒ d x d t = - sin t + sec 2 t 2 2 tan t 2 ⇒ d x d t = - sin t + 1 sin t = cos 2 t sin t ⇒ d y d t = cos t Differentiating with respect to t , we get d 2 y d t 2 = - sint ⇒ d 2 y d t 2 t = π 4 = - 1 2 And, d y d x = d y d t d x d t = cos t × sin t cos 2 t ⇒ d y d x = tan t Again differentiating with respect to x d 2 y d x 2 = sec 2 t × d t d x ⇒ d 2 y d x 2 = s e c 2 t × sin t cos 2 t ∴ d 2 y d x 2 t = π 4 = 2 × 1 2 1 2 2 = 2 2

If x = a sin t and y = a cos t + log tan t 2 , find d 2 y d x 2 .

We have, x = a sin t and y = a cos t + log tan t 2 On differentiating with respect to t we get d x d t = d d t ( a sin t ) = a cos t and, d y d t = d d t a cos t + log tan t 2 = a - sin t + 1 tan t 2 × sec 2 t 2 × 1 2 = a - sin t + 1 2 sin t 2 cos t 2 = a - sin t + 1 sin t ∵ sin 2 A = 2 sin A cos A = a - sin 2 t + 1 sin t = a cos 2 t sin t ∵ cos 2 A + sin 2 A = 1 Now, d y d x = d y d t d x d t = a cos 2 t sin t a cos t = cot ( t ) Therefore, d 2 y d x 2 = d d x d y d x = d d x ( cot ( t ) ) = d d t ( cot ( t ) ) × d t d x = - cosec 2 t × 1 a cos t = - 1 a sin 2 t cos t Hence, d 2 y d x 2 = - 1 a sin 2 t cos t

If x = a ( cos t + t sin t ) and y = a ( sin t - t cos t ) , then find the value of d 2 y d x 2 at t = π 4 .

We have, x = a ( cos t + t sin t ) and y = a ( sin t - t cos t ) On differentiating with respect to t , we get d x dt = d dt [ a ( cos t + t sin t ) ] = - a sin t + a sin t + a t cos t = a t cos t Similarly, d y dt = d dt [ a ( sin t - t cos t ) ] = a cos t - a cos t + a t sin t ⇒ d y d t = a t sin t Now, d y d x = d y dt d x d t = a t sin t a t cos t = tan t d 2 y d x 2 = d d x d y d x = d d x ( tan t ) = d d t ( tan t ) × d t d x = sec 2 t × 1 a t cos t = 1 a t cos 3 t d 2 y d x 2 t = π 4 = 1 a π 4 cos 3 π 4 = 8 2 a π Hence at t = π 4 , d 2 y d x 2 = 8 2 a π

If x = a cos t + log tan t 2 and y = a ( sin t ) , evaluate d 2 y d x 2 at t = π 3 .

We have, x = a cos t + log tan t 2 and y = a sin t On differentiating with respect to t , we get d x d t = d d t a cos t + log tan t 2 = a - sin t + 1 tan t 2 × d d t tan t 2 ⇒ d x d t = a - sin t + 1 tan t 2 × sec 2 t 2 × 1 2 = a - sin t + 1 2 sin t 2 cos t 2 = a - sin t + 1 sin t ∵ sin 2 A = 2 sin A cos A = a - sin 2 t + 1 sin t = a cos 2 t sin t ∵ cos 2 A + sin 2 A = 1 and, d y d t = d d t ( a sin t ) = a cos t Now , d y d x = d y dt dx d t = a cos t a cos 2 t sin t = tan t Therefore, d 2 y d x 2 = d d x d y d x = d d x ( tan ( t ) ) = d d t ( tan ( t ) ) × d t d x = sec 2 t × sin t a cos 2 t = sin t a cos 4 t d 2 y d x 2 t = π 3 = sin π 3 a cos 4 π 3 = 3 2 a 1 16 = 8 3 a Hence, at t = π 3 , d 2 y d x 2 = 8 3 a

If x = a ( cos 2 t + 2 t sin 2 t ) and y = a ( sin 2 t - 2 t cos 2 t ) , then find d 2 y d x 2

We have, x = a ( cos 2 t + 2 t sin 2 t ) and y = a ( sin 2 t - 2 t cos 2 t ) On differentiating with respect to t , we get d x dt = d d t [ a ( cos 2 t + 2 t sin 2 t ) ] = a ( - 2 sin 2 t + 2 sin 2 t + 4 t cos 2 t ) = a ( 4 t cos 2 t ) d y d t = d d t [ a ( sin 2 t - 2 t cos 2 t ) ] = a ( 2 cos 2 t - 2 cos 2 t + 4 t sin 2 t ) = a ( 4 t sin 2 t ) Now, d y d x = d y d t d x d t = a ( 4 t sin 2 t ) a ( 4 t cos 2 t ) = tan 2 t Therefore, d 2 y d x 2 = d d x d y d x = d dx ( tan 2 t ) = d d t ( tan 2 t ) × d t d x = 2 sec 2 2 t × 1 a ( 4 t cos 2 t ) = 1 2 a t sec 3 2 t Hence, d 2 y d x 2 = 1 2 a t sec 3 2 t

If x = 3 cos t - 2 cos 3 t , y = 3 sin t - 2 sin 3 t , find d 2 y d x 2

We have, x = 3 cos t - 2 cos 3 t ⇒ d x d t = 3 ( - sin t ) - 6 cos 2 t ( - sin t ) = - 3 sin t + 6 sin t cos 2 t Also, y = 3 sin t - 2 sin 3 t ⇒ d y d t = 3 cos t - 6 sin 2 t cos t Now, d y d x = d y d t d x d t = 3 cos t - 6 sin 2 t cos t - 3 sin t + 6 sin t cos 2 t = 3 cos t 1 - 2 sin 2 t 3 sin t - 1 + 2 cos 2 t ∵ cos 2 A = 1 - 2 sin 2 A = 2 cos 2 A - 1 = cot t ( cos 2 t ) ( cos 2 t ) = cot t So,, d 2 y d x 2 = d d x d y d x = d d x ( cot t ) = - cosec 2 t · d t d x = - cosec 2 t d x d t = - cosec 2 t - 3 sin t + 6 sin t cos 2 t = - cosec 2 t - 3 sin t 1 - 2 cos 2 t = cosec 2 t - 3 sin t 2 cos 2 t - 1 = cosec 3 t - 3 cos 2 t ∵ cos 2 A = 2 cos 2 A - 1

Find A and B so that y = A sin 3 x + B cos 3 x satisfies the equation d 2 y d x 2 + 4 d y d x + 3 y = 10 cos 3 x

We have, y = A sin 3 x + B cos 3 x Differentiating w.r.t x , we get d y d x = 3 A cos 3 x - 3 B sin 3 x Differentiating w.r.t x , we get d 2 y d x 2 = - 9 A sin 3 x - 9 B cos 3 x Therefore, d 2 y d x 2 + 4 d y d x + 3 y = - 9 A sin 3 x - 9 B cos 3 x + 12 A cos 3 x - 12 B sin 3 x + 3 A sin 3 x + 3 B cos 3 x = ( - 6 A - 12 B ) sin 3 x + ( - 6 B + 12 A ) cos 3 x It is given that, d 2 y d x 2 + 4 d y d x + 3 y = 10 cos 3 x Comparing the coefficients of sin 3 x and cos 3 x , we get - 6 A - 12 B = 0 and - 6 B + 12 A = 10 On solving the equations, we get A = 2 3 and B = - 1 3 Hence, A = 2 3 and B = - 1 3

E M B I B E

MATHEMATICS CLASS XII VOLUME-1>Chapter 12 - Higher Order Derivatives>EXERCISE

R D Sharma Solutions for Chapter: Higher Order Derivatives, Exercise 1: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Higher Order Derivatives, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 12: Higher Order Derivatives, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-1 solutions are prepared by Experienced Embibe Experts.

Questions from R D Sharma Solutions for Chapter: Higher Order Derivatives, Exercise 1: EXERCISE with Hints & Solutions

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 12 - Higher Order Derivatives>EXERCISE>Q 37

If $x = 4 z^{2} + 5, y = 6 z^{2} + 7 z + 3, find \frac{d^{2} y}{d x^{2}}$

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 12 - Higher Order Derivatives>EXERCISE>Q 43

If $x = \cos t + \log (\tan \frac{t}{2}), y = \sin t$ , then find the value of $\frac{d^{2} y}{d t^{2}}$ and $\frac{d^{2} y}{d x^{2}}$ at $t = \frac{π}{4}$ .