Find :

Solution:

The given relationship is $\sin ^{2} y+\cos x y=\pi$

Differentiating this relationship with respect to x, we obtain

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

$\Rightarrow \frac{d}{d x}\left(\sin ^{2} y\right)+\frac{d}{d x}(\cos x y)=0$   ...(1)

Using chain rule, we obtain

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

$\frac{d}{d x}(\cos x y)=-\sin x y \frac{d}{d x}(x y)=-\sin x y\left[y \frac{d}{d x}(x)+x \frac{d y}{d x}\right]$

$=-\sin x y\left[y .1+x \frac{d y}{d x}\right]=-y \sin x y-x \sin x y \frac{d y}{d x}$    ...(3)

From (1), (2), and (3), we obtain

$2 \sin y \cos y \frac{d y}{d x}-y \sin x y-x \sin x y \frac{d y}{d x}=0$

$\Rightarrow(2 \sin y \cos y-x \sin x y) \frac{d y}{d x}=y \sin x y$

$\Rightarrow(\sin 2 y-x \sin x y) \frac{d y}{d x}=y \sin x y$

$\therefore \frac{d y}{d x}=\frac{y \sin x y}{\sin 2 y-x \sin x y}$