#### Answer

For $y\cos x=x^2+y^2$ find $\frac{dy}{dx}$ by implicit differentiation
$\frac{dy}{dx}=\frac{2x +y\sin x}{\cos x-2y}$

#### Work Step by Step

Differentiate both sides with respect to $x$
$\frac{dy}{dx}\cos x-y\sin x=2x+2y\frac{dy}{dx}$
Isolate $\frac{dy}{dx}$
$\frac{dy}{dx}\cos x-2y\frac{dy}{dx}=2x +y\sin x$
$\frac{dy}{dx}(\cos x-2y)=2x +y\sin x$
$\frac{dy}{dx}=\frac{2x +y\sin x}{\cos x-2y}$