Proof for a triple composition chain rule Assuming that $f,g,h$ are all differentiable throughout $\mathbb{R}$, what is the chain rule for $(h \circ g \circ f) (x)$ and the proof for it?
I know that first $(g \circ f)'(x) = g'(f(x))f'(x)$. Is it just applying this rule twice? Then how would one state a proof for such a formula?

You can use associativity:
$$
(h \circ g \circ f)'(x)=(h \circ (g  \circ f))'(x)=h'((g \circ f)(x))\cdot (g\circ f)'(x)
$$
Now $(g \circ f)'(x)=g'(f(x))\cdot f'(x)$ and therefore
$$
(h \circ g \circ f)'(x)=h'((g \circ f)(x))\cdot g'(f(x))\cdot f'(x)
$$
