Suppose f: R -> R has derivatives of all orders. Prove that F(x) := exp(f(x)) also has derivatives of all orders.
The Attempt at a Solution
I can kind of see that this is true but am unsure about how to lay out a proof.
Using the chain rule we get F'(x) = exp'(f(x))f'(x) = exp(f(x))f'(x)
Using the product rule and the chain rule we can again differentiate this as many times as we like. So does my proof need to use induction? If so, how?
Or could I use the Leibnitz formula, if g and h are n-times differntiable then fg is n times differentiable, with g=exp(f(x)) and h=f'(x)? We know h has derivatives of all orders, as does exp. However, what I'm trying to prove is that exp(f(x)) has derivatives of all orders so I can't just claim that g does.