If f^{}(x)= ² x and y=f≤ft({2 x-1}{x²+1}), then d y/d x at x=1 is

Correct Answer is : 12sin⁡2(12)1/2 ²≤ft(1/2)21sin2(21)

Correct Answer is : 12sin⁡2(12)1/2 ²≤ft(1/2)21sin2(21)

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MHTCET 18 Apr 2026 Shift 2 Paper

Question : 123 of 150

Marks: +1, -0

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

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

\frac{d y}{d x}

x=1

Solution:

Concept:

This problem uses the chain rule to differentiate a composite function

y=f(u)

, where

u

is a rational function of

x

Explanation:

Let

u=\frac{2x-1}{x^{2}+1}

. Then

y=f(u)

By the chain rule,

\frac{dy}{dx}=f'(u)\cdot \frac{du}{dx}

Given

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

, so

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

First find

u

x=1

u=\frac{2(1)-1}{1^{2}+1}=\frac{1}{2}

Next, differentiate

u

using the quotient rule:

\frac{du}{dx}=\frac{(x^{2}+1)(2)-(2x-1)(2x)}{(x^{2}+1)^{2}}

x=1

, the numerator is

(2)(2)-(1)(2)=4-2=2

, and the denominator is

(2)^{2}=4

\left.\frac{du}{dx}\right|_{x=1}=\frac{2}{4}=\frac{1}{2}

Thus,

\left.\frac{dy}{dx}\right|_{x=1}=\sin^{2}\left(\frac{1}{2}\right)\cdot \frac{1}{2}

This simplifies to

\frac{1}{2}\sin^{2}\left(\frac{1}{2}\right)

Answer:

\frac{dy}{dx}

x=1

\frac{1}{2}\sin^{2}\left(\frac{1}{2}\right)

, which is Option D.

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