Prove that the directional derivative is the dot product of the gradient and the vector.

Question

I looked for few proofs online but was looking for alternate, more direct proofs. The one on Khan Academy used the Linear Approximation and one used the chain rule of multivariable functions. Are there any alternate methods to prove it? I'm looking for one which doesn't use much except the basic definition of the partial Derivative.

Answer 1

The property holds for differentiable functions, indeed by definition of differentiability we have that

$$\lim_{\vec h\to \vec 0} \frac{ f(\vec x_0+\vec h)-f(\vec x_0)-\nabla f(\vec x_0)\cdot \vec h}{\| \vec h\|}=0 \iff f(\vec x_0+\vec h)-f(\vec x_0)=\nabla f(\vec x_0)\cdot \vec h+o(\| \vec h\|)$$

and assuming $\vec h = t\,\vec v$ we have

$$\frac{\partial f}{\partial \vec v}(\vec x_0)=\lim_{t\to 0}\frac{f(\vec x_0+t\vec v)-f(\vec x_0)}{t}=\lim_{t\to 0}\frac{\nabla f(\vec x_0)\cdot t\vec v+o(\|t\vec v\|)}{t}=\nabla f(\vec x_0)\cdot \vec v$$