Proof of $\frac{\partial f}{\partial\vec u}(x_0,y_0) = ∇f(x_0,y_0)\cdot \vec u $

Question

In order to prove that:

$$\frac{\partial f}{\partial\vec u}(x_0,y_0) = ∇ f(x_0,y_0)\cdot \vec u $$

my book defines:

$$g(t) = f(x_0+at, y_0+bt)$$

then, by the chain rule:

$$g'(0) = \frac{\partial f}{\partial (x_0+at)}\frac{\partial (x_0+at)}{\partial t}+\frac{\partial f}{\partial (y_0+bt)}\frac{\partial (y_0+bt)}{\partial t}$$

which is just:

$$g'(0) = \frac{f(x_0,y_0)}{\partial x}a + \frac{f(x_0,y_0)}{\partial y}b $$

but then the exercise says:

$$g'(0) = \frac{df}{d\vec u}(x_0,y_0)$$

why this?

Answer 1

In general, for a vector $u=(u_1,u_2)$ of norm one, $$ \frac{\partial g}{\partial u}(a,b) = \lim_{\epsilon \to 0} \frac{g(a+\epsilon u_1,b+\epsilon u_2)}{\epsilon}=\lim_{\epsilon \to 0} \frac{g((a,b)+\epsilon u)}{\epsilon} $$

Answer 2

This is by definition of the partial derivative of $f$ in direction $u$ at point $(x_0,y_0)$.