Proof for a triple composition chain rule

Question

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?

Answer 1

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 $$ (h \circ g \circ f)'(x)=h'((g \circ f)(x))\cdot g'(f(x))\cdot f'(x) $$