Method of DifferentiationmediumFree

Method of Differentiation: Question

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

Question

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

and

f'(x)=\sin x^{2}

, then

y'(0)

\sin 1

\cos 1

2\sin 1

correct

\dfrac{\cos 1}{2}

Solution

Step 1: Apply the chain rule:

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

\dfrac{2x-1}{x^{2}+1}

using the quotient rule:

\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 x^{2}

, we have

f'\!\left(\dfrac{2x-1}{x^{2}+1}\right)=\sin\!\left(\dfrac{2x-1}{x^{2}+1}\right)^{2}

. Step 4: At

x=0

\dfrac{2(0)-1}{0+1}=-1

and

\dfrac{-0+0+2}{1}=2

y'(0)=\sin\!\left((-1)^{2}\right)\cdot 2=\sin(1)\cdot 2=2\sin 1

Correct answer: (3)

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