Proof of the second definition of directional derivatives

Suppose \(z=f(x,y)\) is a 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:

$$D_\textbf{u} f(a,b) = \lim_{h \to 0} \frac{f(a+h \cos θ,b+h \sin θ) - f(a,b)}{h}$$

Provided that the limit exist.

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:

$$\begin{align} g'(h) &= \frac{∂f}{∂x}\frac{dx}{dh} + \frac{∂f}{∂y}\frac{dy}{dh} \\ &= f_x (x,y) \cos θ + f_y (x,y) \sin θ \end{align}$$

If \(h=0\), then \(x=a\) and \(y=b\), so:

$$ g'(0) = f_x (a,b) \cos θ + f_y (a,b) \sin θ $$

By definition of \(g'(h)\):

$$\begin{align} g'(0) &= \lim_{h \to 0} \frac{g(h) - g(0)}{h} \\ &= \lim_{h \to 0} \frac{f(a+h \cos θ,b+h \sin θ) - f(a,b)}{h} \end{align}$$

Putting the two together:

$$\begin{align} \lim_{h \to 0} \frac{f(a+h \cos θ,b+h \sin θ) - f(a,b)}{h} &= f_x (a,b) \cos θ + f_y (a,b) \sin θ\\ D_\textbf{u} f(a,b) &= f_x (a,b) \cos θ + f_y (a,b) \sin θ \end{align}$$

Proof Of The Second Definition Of Directional Derivatives

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