Question

Find $\frac{dy}{dx}$ of ${\sin }^{2}x+{\cos }^{2}y=1$

Answer 1

We have ${\sin }^{2}x+{\cos }^{2}y=1$
Differentiating both sides w.r.t. $x,$ we obtain
$\frac{d}{dx}({\sin }^{2}x+{\cos }^{2}y)=\frac{d}{dx}\left(1\right)$
$\Rightarrow 2\sin x.\frac{d}{dx}\left(\sin x\right)+2\cos y.\frac{d}{dx}\left(\cos y\right)=0$
$\Rightarrow 2\sin x\cos x+2\cos y(-\sin y).\frac{dy}{dx}=0$
$\Rightarrow \sin 2x-\sin 2y\frac{dy}{dx}=0$
$\therefore \frac{dy}{dx}=\frac{\sin 2x}{\sin 2y}$