Higher derivatives of exp(f(x))

  • Thread starter Kate2010
  • Start date
  • #1
146
0

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.
 

Answers and Replies

Related Threads on Higher derivatives of exp(f(x))

  • Last Post
Replies
3
Views
2K
  • Last Post
2
Replies
42
Views
11K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
7
Views
9K
  • Last Post
Replies
4
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
Replies
2
Views
1K
Replies
1
Views
796
Replies
3
Views
1K
Top