hey- how do i find a power series for f'(x) if

f(x)= sum from n=0 to n=infinity [(-1^n)*x^(2n)]/[(2n)!]

thanks.

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- Feb 27th 2006, 07:55 PMilovecalculushelp please?
hey- how do i find a power series for f'(x) if

f(x)= sum from n=0 to n=infinity [(-1^n)*x^(2n)]/[(2n)!]

thanks. - Feb 27th 2006, 08:58 PMCaptainBlackQuote:

Originally Posted by**ilovecalculus**

$\displaystyle

f(x)=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}

$?

Differentiate term by term (I know this works for this series so I won't worry

about convergence).

$\displaystyle

f'(x)=\sum_{n=0}^{\infty} (-1)^n \frac{\frac{d(x^{2n})}{dx}}{(2n)!}

$,

so:

$\displaystyle

f'(x)=\sum_{n=0}^{\infty} (-1)^n \frac{2n\ x^{2n-1}}{(2n)!}

$,

Simplifying slightly:

$\displaystyle

f'(x)=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n-1}}{(2n-1)!}

$.

RonL