Q:

Why does $[f(ax)]'=a*f'(ax)$? That is, what effect does the horizontal dilation have on the derivative?

Today my Calculus teacher told us that:
$[f(ax)]'=a*f'(ax)$
This fact was unfortunately not proven, and I do not understand why it is so. It is clear to me that the graph of $y =f(ax)$ will be a transformation of the graph of $y=f(x)$. Specifically, the former will be horizontally dilated by a factor of $1/a$. However, I am not sure what relevance, if any, this transformational reasoning has to the explanation for the statement given at the beginning of this post.
Nor am I sure how (or even if) I can prove this equation using the formal limit definition of the derivative (limit of difference quotient). 

A:

If $a\ne0$, then by substitution $$\lim_{h\to 0}\frac{f(a(x+h))-f(ax)}{h}=\lim_{h\to 0}\frac{f(ax+ah)-f(ax)}{\frac1a\cdot ah}\stackrel{t:=ah\to 0}=\lim_{t\to 0}\frac{f(ax+t)-f(ax)}{\frac1a\cdot t}=\\=af'(ax)$$
If $a=0$ then $x\mapsto f(ax)$ is constant, therefore its derivative is in fact $0$.
As for the geometric interpretation, a horizontal dilation by a factor $\frac1a$ means that all right triangles with catheti parallel to the axes retain the $y$-component cathetus, while having the $x$-component cathetus enlarged by a factor $\frac1a$. This result in slopes changing by a factor $\frac1{1/a}=a$.

