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 \(e^{y}\cos x\neq 0\), \[ \frac{dy}{dx}=-\frac{F_{x}}{F_{y}}=-\frac{-e^{y}\sin x-1}{e^{y}\cos x}=\frac{e^{y}\sin x+1}{e^{y}\cos x} \]