# Prove function $g$ is infinitely differentiable for composite function

Suppose the function $$f$$ is infinitely differentiable on $$\mathbb{R}$$, and the function $$g$$ is differentiable on the open interval $$(a,b)$$ satisfying $$g'(x)=f \circ g \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space x\in(a,b)$$ Prove that the function $$g$$ is infinitely differentiable on $$(a,b)$$.

I have showed that since both functions are differentiable on $$(a,b)$$ implies that they are continuous on $$(a,b)$$, thus $$g'(x)$$ is continuous. Besides that, $$g''(x)=f' \circ g \cdot g'$$ is also continuous. Of course, this process can be done infinitely and hence show that $$g^{(n)}(x)$$ is continuous. How do I show it is infinitely differentiable, because continuous does not imply differentiable.

• Use induction on $n$ to show that the n-th derivative exists for each $n$. Dec 14 '18 at 9:45
• So I have showed $g''$ exists so the function $g$ is twice differentiable on my post? Just use induction to derive the case to $n$-derivative? Dec 14 '18 at 9:47
• For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite product of the functions $f,f',\cdots,f^{(n-1)},g,g',\cdots, g^{(n-1)}$. Dec 14 '18 at 9:52

For a precise argument prove the following statement by induction: for every $$n$$ $$g^{(n)}$$ exists and it is a finite sum of finite products of the functions $$f,f',\cdots,f^{(n-1)},g,g',\cdots, g^{(n-1)}$$.
You know that $$g'(x) = f(g)$$ is differentiable on $$(a,b)$$ by the chain rule ($$f$$ is differentiable on all reals hence on $$(g(a),g(b))$$, and g is differentiable on (a,b)). With the product rule, this implies that $$g''= f '(g) * g'$$ is differentiable, since both f '(g), g' are differentiable. Similarly, $$g''' = (f '(g) )' * g' + (g')' * f '(g) = f ''(g)*(g')^2+ g'' * f '(g)$$ is differentiable since it's a sum of products of differentiable functions. This pattern continues.