Question

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

Answer 1

We have, ${\sin }^{2}y+\cos xy=k$
Differentiating both sides with respect to $x$, we obtain
$\Rightarrow \frac{d}{dx}\left({\sin }^{2}y\right)+\frac{d}{dx}\left(\cos xy\right)=\frac{d\left(\pi \right)}{dx}=0$ .....(1)
Using chain rule, we obtain
$\frac{d}{dx}\left({\sin }^{2}y\right)=2\sin y\frac{d}{dx}\left(\sin y\right)=2\sin y\cos y\frac{dy}{dx}$...... (2)
and

$\frac{d}{dx}\left(\cos xy\right)=-\sin xy\frac{d}{dx}\left(xy\right)=-\sin xy\left[y\frac{d}{dx}\right(x)+x\frac{dy}{dx}]$
$=-\sin xy[y.1+x\frac{dy}{dx}]=-y\sin xy-x\sin xy\frac{dy}{dx}$.....(3)
From (1), (2) and (3), we obtain
$2\sin y\cos y\frac{dy}{dx}-y\sin xy-x\sin xy\frac{dy}{dx}=0$
$\Rightarrow (2\sin y\cos y-x\sin xy)\frac{dy}{dx}=y\sin xy$
$\Rightarrow (\sin 2y-x\sin xy)\frac{dy}{dx}=y\sin xy$
$\therefore \frac{dy}{dx}=\frac{y\sin xy}{\sin 2y-x\sin xy}$