Question
Find $\frac{d y}{d x}$ if $\sin^2 y + \cos xy$ = $\kappa$

Answer

It is given that $\sin^2 y + \cos xy$ = $\kappa$
Differentiating both sides w.r.t. $x$, we get,
$\frac{d}{d x}\left(\sin ^{2} y+\cos x y\right)=\frac{d}{d x}(\kappa)$
$\Rightarrow 2 \sin y \cos y \frac{d y}{d x}-\sin x y\left[y \frac{d}{d x}(x)+x \frac{d y}{d x}\right]=0$
$\Rightarrow 2 \sin y \cos y \frac{d y}{d x}-\sin x y\left[y \cdot 1+x \frac{d y}{d x}\right]=0$
$\Rightarrow 2 \sin y \cos y \frac{d y}{d x}-y \sin x y-x \sin x y \frac{d y}{d x}=0$
$\Rightarrow(2 \sin y \cos y-x \sin x y) \frac{d y}{d x}=y \sin x y$
$\Rightarrow(\sin 2 y-x \sin x y) \frac{d y}{d x}=y \sin x y$
$\Rightarrow \frac{d y}{d x}=\frac{\text { y sin xy }}{(\sin 2 y-x \sin x y)}$

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