Implicit domain how to prove dy/dx = -(sqrt(1-y^2))/(sqrt(1-x^2) ?

If $x \sqrt{1 - {y}^{2}} + y \sqrt{1 - {x}^{2}} = 0$ Show that $\frac{\mathrm{dy}}{\mathrm{dx}}$ = -(sqrt(1-y^2))/(sqrt(1-x^2)

Apr 28, 2018

Explanation:

Here,

$x \sqrt{1 - {y}^{2}} + y \sqrt{1 - {x}^{2}} = 0. . . \to \left(A\right)$

Let,

x=sinalpha=>alpha=sin^-1x,where,alphain(-pi/4,pi/4);and

$y = \sin \beta \implies \beta = {\sin}^{-} 1 y , w h e r e . \beta \in \left(- \frac{\pi}{4} , \frac{\pi}{4}\right) .$

$\implies \alpha + \beta \in \left(\left(- \frac{\pi}{4}\right) + \left(- \frac{\pi}{4}\right) , \frac{\pi}{4} + \frac{\pi}{4}\right)$

$\implies \alpha + \beta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

So, from $\left(A\right)$

$\sin \alpha \sqrt{1 - {\sin}^{2} \beta} + \sin \beta \sqrt{1 - {\sin}^{2} \alpha} = 0$

$\implies \sin \alpha \cos \beta + \sin \beta \cos \alpha = 0$

$\implies \sin \left(\alpha + \beta\right) = 0$

$\implies \alpha + \beta = {\sin}^{-} 1 \left(0\right)$

$\implies \alpha + \beta = 0$

Subst. back, $\alpha \mathmr{and} \beta$

$\implies {\sin}^{-} 1 x + {\sin}^{-} 1 y = 0$

Diff.w.r.t.$x$,

$\frac{1}{\sqrt{1 - {x}^{2}}} + \frac{1}{\sqrt{1 - {y}^{2}}} \times \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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

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

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