Find $\dfrac{dy}{dx}$ if $y=f(x)$ is defined implicitly by $F(x,y)=e^{y}\cos x-x-1=0$.

Solution First we find the partial derivatives of $F$. $$F_{x}=\dfrac{\partial F}{\partial x}=-e^{y}\sin x-1\qquad \hbox{and} \qquad F_{y}=\dfrac{ \partial F}{\partial y}=e^{y}\cos x$$

Then we use (3). If $e^{y}\cos x\neq 0$, $$\frac{dy}{dx}=-\frac{F_{x}}{F_{y}}=-\frac{-e^{y}\sin x-1}{e^{y}\cos x}=\frac{e^{y}\sin x+1}{e^{y}\cos x}$$