Suppose \(z=f(x,y)\) is a differentiable function of two variables with a domain of \(D\). Given a point \((a,b)\) in the domain of \(f\), we choose a direction to travel from that point.
We measure the direction using an angle \(θ\), which is measured counterclockwise in the \(x\), \(y\)-plane, starting at zero from the positive \(x\)-axis. The distance we travel is \(h\) (where \(h\) is parallel to the \(x\), \(y\)-plane) and the direction we travel is given by the unit vector \(\textbf{u}=(\cos θ)\textbf{i}+(\sin θ)\textbf{j}\).
The directional derivative of \(f\) in the direction of \(\textbf{u}\) is given by:
Provided that the limit exist.
We will give two proofs. The first proof involves the fact the \(f\) is differentiable. This means if you zoom in at the surface of \(f\) very close to \((a,b)\), the surface would be look more like the tangent plane at \((a,b)\). If you go even more close, the surface would look almost exactly like the tangent plane.
The derivative of the tangent plane in the \(y\) direction is \(f_y(a,b)\), the derivative in the \(x\) direction is \(f_x(a,b)\), and the derivative in the \(\textbf{u}\) direction is \( [D_{\textbf{u}}f(a,b) = f_x (a,b) \cos θ + f_y (a,b) \sin θ] \) (proof here).
This is also the directional drivative of the surface, since the surface is suppose to be approaching the tangent plane based on the definition of differentiability.
As for the second proof, let \([x = a + h \cos θ]\) and \([y =b+h \sin θ]\), and define \(g(h)=f(x,y)\). Since \(f_x\) and \(f_y\) both exist, and therefore \(f\) is differentiable, that means:
If \(h=0\), then \(x=a\) and \(y=b\), so:
By definition of \(g'(h)\):
Putting the two together: