Prove Product rule by Caratheodory’s Theorem

Question

Let $I$ be an interval, let $c\in I$ and let $f,g : I \to \mathbb{R}$ be differentiable at $c$. By Caratheodory’s theorem there exist functions $φ,ψ : I \to \mathbb{R}$ which are continuous at $c$ and with $φ(c) = f′(c)$ and $ψ(c) = g′(c)$ such that for all $x \in I$ we have $$f(x) = f(c) + φ(x)(x − c), g(x) = g(c) + ψ(x)(x − c).$$

Use these representations to prove the product rule i.e. prove that $fg$ is differentiable at $c$ and that $$(fg)′(c) = f′(c)g(c) + f(c)g′(c).$$

Answer 1

You have $$f(x)g(x)=f(c)g(c)+\biggl(f(c)\psi(x)+g(c)\phi(x)+\phi(x)\psi(x)(x-c)\biggr)(x-c)\ .$$ Now analyze the large parenthesis when $x\to c$, and appeal to the other direction of Caratheodory's theorem.