Question

If $y\sqrt{1-x^2}+x\sqrt{1-y^2}=1$, prove that $\frac{dy}{dx}=-\sqrt{\frac{1-y^2}{1-x^2}}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y\sqrt{1-x^2}+x\sqrt{1-y^2}=1 \\ \mathrm{Let}x=\sin A,y=\sin B \\ \Rightarrow \sin B\sqrt{1-{\sin }^{2}A}+\sin A\sqrt{1-{\sin }^{2}B}=1 \\ \Rightarrow \sin B\cos A+\sin A\cos B=1\left[\because \sin \left(x+y\right)=\sin x\cos y+\cos x\sin y\right] \\ \Rightarrow \sin \left(A+B\right)=1 \\ \Rightarrow A+B={\sin }^{-1}\left(1\right) \\ \Rightarrow {\sin }^{-1}x+{\sin }^{-1}y=\frac{\mathrm{\pi }}{2}\left[\because x=\sin A,y=\sin B\right] \end{gathered}$$
Differentiate with respect to x,
$$\begin{gathered} \Rightarrow \frac{d}{dx}\left({\sin }^{-1}x\right)+\frac{d}{dx}\left({\sin }^{-1}y\right)=\frac{d}{dx}\left(\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-y^2}}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=0 \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-\sqrt{\frac{1-y^2}{1-x^2}} \end{gathered}$$