JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

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

, then

\frac{{dy}}{{dx}}

equals

View answer and explanation

Correct option: A

2x[1 + \tan (\log x)] + x{\sec ^2}(\log x)

Option A (

2x[1 + \tan (\log x)] + x{\sec ^2}(\log x)

) is correct because it is the result of differentiating the given equation and solving for dy/dx.

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x

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