TITLE: Proving the chain rule by first principles
QUESTION [5 upvotes]: I'm currently trying to prove:
$(f(g))'(a)=f'(g(a))*g'(a)$
I have been given a proof which manipulates:
$f(a+h)=f(a)+f'(a)h+O(h)$ where $O(h)$ is the error function. However, I would like to have a proof in terms of the standard limit definition of $(1/h)*(f(a+h)-f(a) \to f'(a)$ as $h \to 0$
Thanks!

REPLY [11 votes]: Since $g$ is differentialiable at the point $a$ then it'z continuous and then 
$$\lim_{x\to a}g(x)=g(a)$$ 
hence
$$(f\circ g)'(a)=\lim_{x\to a}\frac{f(g(x))-f(g(a))}{x-a}=\lim_{x\to a}\frac{f(g(x))-f(g(a))}{g(x)-g(a)}\frac{g(x)-g(a)}{x-a}\\=\lim_{y\to g(a)}\frac{f(y)-f(g(a))}{y-g(a)}\lim_{x\to a}\frac{g(x)-g(a)}{x-a}=f'(g(a))g'(a)$$