# The derivative of a Taylor series?

1. Jun 27, 2012

I took my first calculus class over the last two semesters, and my teacher and I privately worked on some harder material together. Toward the end of the school year he gave me a question that I never answered and never found an answer for. It asked me to find the derivative of a Taylor series. I don't remember the specific problem, but if someone could shed some light, that'd be lovely!

2. Jun 27, 2012

### mathwonk

you are kidding right? i.e. what do you think?

3. Jun 27, 2012

I'm not kidding at all. At the time I received the problem I'd never heard of a Taylor series, and I haven't seen a problem with one since. It's something I've never had the opportunity to learn at school thus far.

4. Jun 27, 2012

### A. Bahat

I'm not sure if this is exactly what you're looking for, but if f(x)=Ʃanxn in some interval, then f'(x)=Ʃnanxn-1 in the interior of that interval; that is, you can differentiate the series termwise. This is because of uniform convergence, which can be shown with the Weierstrass M-test.

5. Jun 27, 2012

### A. Bahat

Really, it's a very plausible generalization of what you already know for finite polynomials.

6. Jun 27, 2012

I'm really just caught up in looking at the sigma notation.. It throws me out of whack. I'm assuming I'm making a much bigger deal out of this than I should be.

This may be a better question:

If f(x) = f(a) + [f'(a)/1!](x-a) + [f"(a)/2!](x-a)^2 + ...
Then f'(x) = ?

Last edited: Jun 27, 2012
7. Jun 27, 2012

### pwsnafu

Exactly what you would get if you differentiated each term individually.
Provided it converges.

8. Jun 27, 2012

### lugita15

badamann, just use the power rule.

9. Jun 28, 2012