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Question

Mathematics Question on Differentiability

If sin⁡−1 x+sin⁡−1 y=π2,\sin^{-1} \, x + \sin^{-1} \, y = \frac{\pi}{2}, then dydx\frac{dy}{dx} is equal to

A

xy\frac{x}{y}

B

−xy- \frac{x}{y}

C

yx\frac{y}{x}

D

−yx- \frac{y}{x}

Answer

−xy- \frac{x}{y}

Explanation

Solution

Given that,
sin⁡−1x+sin⁡−1y=π2\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}
∴sin⁡−1x=cos⁡−1y\therefore \sin ^{-1} x=\cos ^{-1} y
⇒y=1−x2\Rightarrow y=\sqrt{1-x^{2}}
On differentiating with respect to xx,
we get dydx=−2x21−x2=−xy\frac{d y}{d x}=\frac{-2 x}{2 \sqrt{1-x^{2}}}=-\frac{x}{y}