Question

Find $\frac{dy}{dx}$ if ${\sin }^2\left(x\right)+{\cos }^2\left(y\right)=1$.

Answer 1

Compute the required derivative:

$\Rightarrow {\sin }^{2}x+{\cos }^{2}y=1$ (Given)

Differentiating both sides with respect to$x$, we get,

Formula $\left[\because \frac{d}{dx}{\sin }^2\left(x\right)=2\sin \left(x\right)\frac{d}{dx}\sin \left(x\right)=2\sin x·\cos x,\frac{d}{dx}\left(cons\tan t\right)=0\right]$

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

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

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

$\Rightarrow$$\frac{dy}{dx}=\frac{\sin \left(2x\right)}{\sin \left(2y\right)}$

Hence, the correct answer is $\frac{\sin \left(2x\right)}{\sin \left(2y\right)}$.