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JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1
If sin⁡−1(x2−y2x2+y2)=log⁡a{\sin ^{ - 1}}\left( {\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}} \right) = \log a, then dydx\frac{{dy}}{{dx}} equals
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Correct option: D
yx\frac{y}{x}
Option D (yx\frac{y}{x}) is correct because implicit differentiation leads to the relationship dydx=yx\frac{dy}{dx} = \frac{y}{x}.

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