Maclaurin series of $\frac{1}{1+x^2}$

I'm stumped here. I''m supposed to find the Maclaurin series of $\frac1{1+x^2}$, but I'm not sure what to do. I know the general idea: find $\displaystyle\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$. What I want to do is find some derivatives and try to see if there's a pattern to their value at $0$. But after the second derivative or so, it becomes pretty difficult to continue. I know this:

$$f(0) = 1$$

$$f'(0) = 0$$

$$f''(0) = -2$$

$$f^{(3)}(0) = 0$$

$$f^{(4)}(0) = 0$$

But when trying to calculate the fifth derivative, I sort of gave up, because it was becoming too unwieldly, and I didn't even know if I was going somewhere with this, not to mention the high probability of making a mistake while differentiating.

Is there a better of way of doing this? Differentiaing many times and then trying to find a pattern doesn't seem to be working.

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$1$) Write down the series for $\frac{1}{1-t}$. You have probably have already seen this one. If not, it can be computed by the method you were using on $\frac{1}{1+x^2}$. The derivatives are a lot easier to get a handle on than the derivatives of $\frac{1}{1+x^2}$.
$2$) Substitute $-x^2$ for $t$, and simplify.
Comment: It can be quite difficult to find an expression for the $n$-th derivative of a function. In many cases, we obtain the power series for a function by "recycling" known results. In particular, we often get new series by adding known ones, or by differentiating or integrating known ones term by term. Occasionally, substitution is useful.
It is worth taking home the explicit lesson that successive differentiation is not usually a practical way to find power series (except in special lucky cases such as $e^x$). It is more of theoretical importance (such as in proving that the power series for a function is unique if it exists). –  Henning Makholm Oct 28 '11 at 18:26
Thanks, I get it. I left a comment but deleted it because I had made a mistake. I now get $\displaystyle \sum_{n=0}^\infty (-1)^n x^{2n}$. Is that right? –  Javier Badia Oct 28 '11 at 18:29
Yes, that's right. I had left a comment giving the $(-1)^n$ stuff, but deleted it when you deleted your question, since I decided you had figured it out. By the way, soon you will be calculating new series by differentiating or integrating known ones term by term. –  André Nicolas Oct 28 '11 at 18:37