If $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1, \quad$ then $\frac{d y}{d x}=$

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Aicharya · Accepted Answer

Given $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$ Put $x=\sin \alpha$ and $y=\sin \beta$ Given equation becomes $\sin \beta \cos \alpha+\sin \alpha \cos \beta=1 \Rightarrow \sin (\alpha+\beta)=1 \Rightarrow \alpha+\beta=\sin ^{-1}(1)$ $	herefore \sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}$ Differentiating w.r.t. $x$ $\begin{aligned} & \frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-y^{2}} \frac{d y}{d x}}=0 \ 	herefore & \frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}=-\frac{1}{\sqrt{1-x^{2}}} \Rightarrow \frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}} \end{aligned}$

If $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1, \quad$ then $\frac{d y}{d x}=$

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