Question

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

Answer 1

Given: ${\sin }^{2}x+{\cos }^{2}y=1$
Differentiating both sides w.r.t. $x$, we get,

$\frac{d\left({\sin }^{2}x\right)}{dx}+\frac{d\left({\cos }^{2}y\right)}{dx}=\frac{d\left(1\right)}{dx}$

$\Rightarrow \frac{d\left({\sin }^{2}x\right)}{dx}+\frac{d(\cos y{)}^2}{dx}\times \frac{dy}{dx}=0$

$\Rightarrow 2\sin x\frac{d\left(\sin x\right)}{dx}+2\cos y\frac{d\left(\cos y\right)}{dy}\times \frac{dy}{dx}=0$

$\Rightarrow 2\sin x\cos x+2\cos y(-\sin y)\frac{dy}{dx}=0$

$\Rightarrow 2\sin x\cos x=2\cos y\sin y\frac{dy}{dx}$

$\Rightarrow \sin 2x=\sin 2y\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{\sin 2x}{\sin 2y}$