Find dydx if y=f(x) is defined implicitly by F(x,y)=eycosx−x−1=0.
Solution First we find the partial derivatives of F. Fx=∂F∂x=−eysinx−1andFy=∂F∂y=eycosx
Then we use (3). If eycosx≠0, dydx=−FxFy=−−eysinx−1eycosx=eysinx+1eycosx