Higher derivatives of exp(f(x))

[Kate2010]

Homework Statement

Suppose f: R -> R has derivatives of all orders. Prove that F(x) := exp(f(x)) also has derivatives of all orders.

Homework Equations

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.

 

[mighty2000]

it is important to always provide a clear and logical proof for any statement or claim. In this case, we want to prove that the function F(x) := exp(f(x)) has derivatives of all orders.

First, we can start by using the definition of a derivative. We know that the derivative of a function at a point x is defined as the limit of the difference quotient as the change in x approaches 0. In other words, the derivative of F(x) at a point x is given by:

F'(x) = lim (F(x+h) - F(x))/h as h approaches 0

Next, we can use the chain rule to expand F(x+h) and F(x):

F'(x) = lim (exp(f(x+h)) - exp(f(x)))/h as h approaches 0
= lim (exp(f(x+h)) - exp(f(x)))/(f(x+h) - f(x)) * (f(x+h) - f(x))/h as h approaches 0

Now, we know that f(x) is differentiable at x and has derivatives of all orders. This means that (f(x+h) - f(x))/h has derivatives of all orders as well. And since we are taking the limit as h approaches 0, we can assume that this term is well-behaved and has derivatives of all orders.

Additionally, we also know that exp(x) has derivatives of all orders. This means that (exp(f(x+h)) - exp(f(x)))/(f(x+h) - f(x)) also has derivatives of all orders, as it is a composition of two functions with derivatives of all orders.

Hence, we have shown that F'(x) has derivatives of all orders. This means that F(x) = exp(f(x)) has derivatives of all orders as well.

We can now use induction to extend this proof to show that all higher order derivatives of F(x) are also well-defined and have derivatives of all orders. This is because we can use the same argument above to show that F''(x), F'''(x), and so on, all have derivatives of all orders.

Therefore, we have proven that F(x) := exp(f(x)) has derivatives of all orders.