If $\sin (\mathrm{xy})+\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}^{2}-\mathrm{y}^{2}$, find $\frac{\mathrm{dy}}{\mathrm{dx}}$.
We are given with an equation $\sin (x y)+\frac{y}{x}=x^{2}-y^{2}$, we have to find $\frac{d y}{d x}$ by using the given equation, so by
differentiating the equation on both sides with respect to $x$, we get,
$\cos (x y)\left[(1) y+x \frac{d y}{d x}\right]+\frac{x \frac{d y}{d x}-y(1)}{x^{2}}=2 x-2 y \frac{d y}{d x}$
$y \cos (x y)+x \cos (x y) \frac{d y}{d x}+\frac{1}{x} \frac{d y}{d x}-\frac{y}{x^{2}}=2 x-2 y \frac{d y}{d x}$
$\frac{d y}{d x}\left[x \cos (x y)+\frac{1}{x}+2 y\right]=2 x-y \cos (x y)+\frac{y}{x^{2}}$
$\frac{d y}{d x}=\frac{2 x-y \cos (x y)+\frac{y}{x^{2}}}{x \cos (x y)+\frac{1}{x}+2 y}$
$\frac{d y}{d x}=\frac{2 x^{3}-y x^{2} \cos (x y)+y}{x\left[x^{2} \cos (x y)+1+2 x y\right]}$
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