Question

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

Answer 1

$x\sqrt{1-y^2}+y\sqrt{1-x^2}=1$

Differentiating throughout with respect to x we get

$x\frac{1}{2\sqrt{1-y^2}}\cdot (-2y)\frac{dy}{dx}+\sqrt{1-y^2}+y\frac{1}{2\sqrt{1-x^2}}\cdot (-2x)+\sqrt{1-x^2}\frac{dy}{dx}=0$

$\Rightarrow \frac{-xy}{\sqrt{1-y^2}}\frac{dy}{dx}+\sqrt{1-y^2}+\frac{(-xy)}{\sqrt{1-x^2}}+\sqrt{1-x^2}\frac{dy}{dx}=0$

$\Rightarrow (\frac{-xy}{\sqrt{1-y^2}}+\sqrt{1-x^2})\frac{dy}{dx}=\frac{dy}{\sqrt{1-x^2}}-\sqrt{1-y^2}$

$\Rightarrow \left(\frac{-xy+\sqrt{(1-x^2)(1-y^2)}}{\sqrt{1-y^2}}\right)\frac{dy}{dx}=\left(\frac{xy-\sqrt{(1-x^2)(1-y^2)}}{\sqrt{1-x^2}}\right)$

$\Rightarrow \frac{-(xy-\sqrt{(1-x^2)(1-y^2)})}{\sqrt{1-y^2}}$ $\frac{dy}{dx}$ $=\frac{(xy-\sqrt{(1-x^2)(1-y^2)})}{\sqrt{1-x^2}}$

$\Rightarrow \frac{dy}{dx}=-\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$

Hence $\frac{dy}{dx}=-\sqrt{\frac{1-y^2}{1-x^2}}$.