Method of DifferentiationhardFree

Method of Differentiation: Question

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

Question
If f′′(x)=excos⁡xf''(x)=e^{x}\cos x, lim⁡x→−∞f′(x)=0\displaystyle\lim_{x\to-\infty}f'(x)=0 and y=f ⁣(x−1x+1)y=f\!\left(\dfrac{x-1}{x+1}\right), then dydx\dfrac{dy}{dx} at x=1x=1 is
Asin⁡12\dfrac{\sin 1}{2}correct
Bsin⁡1\sin 1
Cesin⁡1e\sin 1
Desin⁡12\dfrac{e\sin 1}{2}
Solution
Step 1: Integrate f′′(x)=excos⁡xf''(x)=e^{x}\cos x to find f′(x)f'(x). Substitute t=ext=e^{x}, so dt=exdxdt=e^{x}dx. Then cos⁡x=cos⁡(ln⁡t)\cos x=\cos(\ln t).
f′(x)=∫excos⁡x dx=∫cos⁡t dt∣t=ex=sin⁡(ex)+Cf'(x)=\int e^{x}\cos x\,dx=\int\cos t\,dt\bigg|_{t=e^{x}}=\sin(e^{x})+C
 Step 2: Apply the condition lim⁡x→−∞f′(x)=0\displaystyle\lim_{x\to-\infty}f'(x)=0. As x→−∞x\to-\infty: ex→0e^{x}\to 0, so sin⁡(ex)→sin⁡0=0\sin(e^{x})\to\sin 0=0. Therefore C=0C=0:
f′(x)=sin⁡(ex)f'(x)=\sin(e^{x})
 Step 3: Differentiate y=f ⁣(x−1x+1)y=f\!\left(\dfrac{x-1}{x+1}\right) using the chain rule:
dydx=f′ ⁣(x−1x+1)⋅ddx ⁣(x−1x+1)\dfrac{dy}{dx}=f'\!\left(\dfrac{x-1}{x+1}\right)\cdot\dfrac{d}{dx}\!\left(\dfrac{x-1}{x+1}\right)
ddx ⁣(x−1x+1)=(x+1)−(x−1)(x+1)2=2(x+1)2\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=1x=1: x−1x+1=0\dfrac{x-1}{x+1}=0 and 2(x+1)2=24=12\dfrac{2}{(x+1)^{2}}=\dfrac{2}{4}=\dfrac{1}{2}:
dydx∣x=1=f′(0)⋅12=sin⁡(e0)⋅12=sin⁡(1)⋅12=sin⁡12\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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