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MHTCET 23 Apr 2025 Shift 2 Paper

Question : 145 of 150

Marks: +1, -0

x=e^{\tan ^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)}

\frac{d y}{d x}

x=1

Solution:

Concept:

Simplify the exponential equation using logarithms, then differentiate using the chain rule.

Explanation:

Given:

x=e^{\tan^{-1}\left(\frac{y-x^2}{x^2}\right)}

Taking natural logarithm on both sides:

\log x=\tan^{-1}\left(\frac{y-x^2}{x^2}\right)

Taking tangent on both sides:

\tan(\log x)=\frac{y-x^2}{x^2}

Rearranging:

y=x^2\tan(\log x)+x^2

Differentiating with respect to

x

\frac{dy}{dx}=2x\tan(\log x)+x^2\cdot\frac{\sec^2(\log x)}{x}+2x

=x\sec^2(\log x)+2x\tan(\log x)+2x

x=1

\log 1=0

\tan 0=0

\sec^2 0=1

\left.\frac{dy}{dx}\right|_{x=1}=1(1)+2(1)(0)+2(1)=3

Answer:

\frac{dy}{dx}=3

x=1

, so the correct option is D.

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