Question:
Find $\frac{\mathrm{dy}}{\mathrm{dx}}$ in each of the following:
$\sin x y+\cos (x+y)=1$
Solution:
We are given with an equation $\sin x y+\cos (x+y)=1$, we have to find $\frac{d y}{d x}$ of it, so by differentiating the equation on both sides with respect to $x$, we get,
$\cos x y\left(y+x \frac{d y}{d x}\right)-\sin (x+y)\left(1+\frac{d y}{d x}\right)=0$
$\frac{d y}{d x}[x \cos x y-\sin (x+y)]=\sin (x+y)-y \cos x y$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sin (\mathrm{x}+\mathrm{y})-\mathrm{y} \cos \mathrm{xy}}{\mathrm{x} \cos \mathrm{xy}-\sin (\mathrm{x}+\mathrm{y})}$