$ x=e^{\tan ^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)} $, then $ \frac{d y}{d x} $ is

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Hard

1 of 1

Application +4 | -1

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

, then

\frac{d y}{d x}

View answer and explanation

Correct option: A

x\left(2+\frac{y^{2}}{x^{4}}-\frac{2 y}{x^{2}}\right)+\frac{2 y}{x}

The correct answer is A (

x\left(2+\frac{y^{2}}{x^{4}}-\frac{2 y}{x^{2}}\right)+\frac{2 y}{x}

) because it is the derivative of y with respect to x after implicit differentiation and simplification.

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