Method of DifferentiationhardFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

f''(x)=e^{x}\cos x

\displaystyle\lim_{x\to-\infty}f'(x)=0

and

y=f\!\left(\dfrac{x-1}{x+1}\right)

, then

\dfrac{dy}{dx}

x=1

\dfrac{\sin 1}{2}

correct

\sin 1

e\sin 1

\dfrac{e\sin 1}{2}

Solution

Step 1: Integrate

f''(x)=e^{x}\cos x

to find

f'(x)

. Substitute

t=e^{x}

, so

dt=e^{x}dx

. Then

\cos x=\cos(\ln t)

f'(x)=\int e^{x}\cos x\,dx=\int\cos t\,dt\bigg|_{t=e^{x}}=\sin(e^{x})+C

Step 2: Apply the condition

\displaystyle\lim_{x\to-\infty}f'(x)=0

. As

x\to-\infty

e^{x}\to 0

, so

\sin(e^{x})\to\sin 0=0

. Therefore

C=0

f'(x)=\sin(e^{x})

Step 3: Differentiate

y=f\!\left(\dfrac{x-1}{x+1}\right)

using the chain rule:

\dfrac{dy}{dx}=f'\!\left(\dfrac{x-1}{x+1}\right)\cdot\dfrac{d}{dx}\!\left(\dfrac{x-1}{x+1}\right)

\dfrac{d}{dx}\!\left(\dfrac{x-1}{x+1}\right)=\dfrac{(x+1)-(x-1)}{(x+1)^{2}}=\dfrac{2}{(x+1)^{2}}

Step 4: At

x=1

\dfrac{x-1}{x+1}=0

and

\dfrac{2}{(x+1)^{2}}=\dfrac{2}{4}=\dfrac{1}{2}

\dfrac{dy}{dx}\bigg|_{x=1}=f'(0)\cdot\dfrac{1}{2}=\sin(e^{0})\cdot\dfrac{1}{2}=\sin(1)\cdot\dfrac{1}{2}=\dfrac{\sin 1}{2}

Correct answer: (1)

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