Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

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

Question
If y=f ⁣(2x−1x2+1)y=f\!\left(\dfrac{2x-1}{x^{2}+1}\right) and f′(x)=sin⁡x2f'(x)=\sin x^{2}, then y′(0)y'(0) is
Asin⁡1\sin 1
Bcos⁡1\cos 1
C2sin⁡12\sin 1correct
Dcos⁡12\dfrac{\cos 1}{2}
Solution
Step 1: Apply the chain rule:
y′(x)=f′ ⁣(2x−1x2+1)⋅ddx ⁣(2x−1x2+1)y'(x)=f'\!\left(\dfrac{2x-1}{x^{2}+1}\right)\cdot\dfrac{d}{dx}\!\left(\dfrac{2x-1}{x^{2}+1}\right)
 Step 2: Differentiate 2x−1x2+1\dfrac{2x-1}{x^{2}+1} using the quotient rule:
ddx ⁣(2x−1x2+1)=2(x2+1)−(2x−1)(2x)(x2+1)2=2x2+2−4x2+2x(x2+1)2=−2x2+2x+2(x2+1)2\dfrac{d}{dx}\!\left(\dfrac{2x-1}{x^{2}+1}\right)=\dfrac{2(x^{2}+1)-(2x-1)(2x)}{(x^{2}+1)^{2}}=\dfrac{2x^{2}+2-4x^{2}+2x}{(x^{2}+1)^{2}}=\dfrac{-2x^{2}+2x+2}{(x^{2}+1)^{2}}
 Step 3: Since f′(x)=sin⁡x2f'(x)=\sin x^{2}, we have f′ ⁣(2x−1x2+1)=sin⁡ ⁣(2x−1x2+1)2f'\!\left(\dfrac{2x-1}{x^{2}+1}\right)=\sin\!\left(\dfrac{2x-1}{x^{2}+1}\right)^{2}. Step 4: At x=0x=0: 2(0)−10+1=−1\dfrac{2(0)-1}{0+1}=-1 and −0+0+21=2\dfrac{-0+0+2}{1}=2:
y′(0)=sin⁡ ⁣((−1)2)⋅2=sin⁡(1)⋅2=2sin⁡1y'(0)=\sin\!\left((-1)^{2}\right)\cdot 2=\sin(1)\cdot 2=2\sin 1
 Correct answer: (3)
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