Find $f^{(1001)}(0)$ I am to find the value in 0 of 1001th derivative of the function $$f(x) = \frac{1}{2+3x^2}$$
How should I approach this kind of problem? I tried something like : $$\frac{1}{2+3x^2} = \frac{1}{2}\cdot\frac{1}{1-(-\frac{3}{2}x^2)}= \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{3x^2}{2}\right)^n$$ and compare whats next to $x^{1001}$ in this sum and in MacLaurin series but damn, we've got only even powers of x here. How should I do that?
 A: Just note that the Taylor series of an even function has only terms of even powers of $x$ and odd powers of $x$ if the function is odd. So, without computing the series you should know the answer. 
So, you need only to see the function is even or odd.
A: You’re missing a minus sign:
$$f(x)=\frac12\sum_{n\ge 0}(-1)^n\left(\frac32\right)^nx^{2n}\;.$$
Thus, 
$$f^{(1001)}(x)=\frac12\sum_{n\ge 0}(-1)^n\left(\frac32\right)^n(2n)^{\underline{1001}}x^{2n-1001}\;,$$
where $$(2n)^{\underline{1001}}=(2n)(2n-1)\ldots(2n-1001+1)=(2n)(2n-1)\ldots(2n-1000)$$ is a falling factorial. This is zero if $n\le 500$, so
$$\begin{align*}
f^{(1001)}(x)&=\frac12\sum_{n\ge 501}(-1)^n\left(\frac32\right)^n(2n)^{\underline{1001}}x^{2n-1001}\\\\
&=\frac12\left(-\left(\frac32\right)^{501}x+\left(\frac32\right)^{502}x^3-\ldots\right)\;,
\end{align*}$$
and $f^{(1001)}(0)=0$.
Of course you don’t actually have to do all of this calculation: the fact that the original series has only even powers means that any odd-order derivative must have only odd powers and hence a constant term of $0$.
A: "we've got only even powers of $x$ here" is exactly what will help you! By examining the Taylor series you can see that the coefficient of the $x^{1001}$ term is $0$. 
Therefore, after taking the derivatives you will have $0 \cdot 1001! x^0$ as your constant term. Since all other terms in the Taylor series expansion have $x^m$ where $m$ is positive, you can conclude that $$f^{(1001)} = 0 + 0 + \cdots 0 = 0.$$ 
