I'm trying to answer Tu's Introduction to Manifolds exercise 1.3, which says in the first item:
Let $U \subset \mathbb{R}^n$ and $V \subset \mathbb{R}^n$ be open subsets. A $C^\infty$ map $F \colon U \to V$ is called a diffeomorphism is it is bijective and has a $C^\infty$ inverse $F^{-1} \colon V \to U$. Show that the function $f \colon ]\frac{-\pi}2, \frac\pi2[ \to \mathbb{R}$ given by $f(x) = \tan x$ is a diffeomorphism.
By doing some calculations, I arrived at the equality in the title: $\frac{d^2f}{dx^2} = f\frac{df}{dx}$. Note that both $f$ and $\frac{df}{dx}$ are continuous and defined everywhere in the given domain. Is it possible to justify formally that this implies that $f$ is in fact $C^\infty$? It is pretty clear to me that this is in fact true, because differentiating both sides gives $\frac{d^3f}{dx^3}$ in function of $\frac{df}{dx}$ and $f$ again, and this keeps happening forever in a way which preserves continuity for all derivatives. The problem is that I really don't know how to justify this formally.
From the product rule, we see that if $g$ and $h$ are differentiable functions, then so it the product $gh$.
Thus the equation $$ f''=ff' $$ also shows that $f''$ is differentiable, since $f$ and $f'$ are. Repeat this logic to show that each derivative $f^{(n)}$ of $f$ exists and is differentiable.