Computing taylor series, getting all 0's 
I started out by finding the first and second derivative.
For $f'(x)$ I got $\;\;\dfrac{(12x^2-x^4)}{(4-x^2)^2}$
For $f''(x)$ I got $\;\;\dfrac{(4-x^2)(24x-4x^3)-(12x^2-x^4)(-4x) }{ (4-x^2)^3}$
After evaluating $f'(0)$ and $f''(0)$ I got $0$ for both of those ($f'(0)=0$ and $f''(0)= 0$), also for $f(0)$ I got $0$.
This is confusing me because if I go to plug the values into the Taylor series function I just get $0 + 0x + \dfrac{(0*x^2)}{(2!)}$.
Is that actually right or did I go wrong somewhere?
 A: You can certainly compute derivatives to find the Taylor Series about $c=0$ for this function. However, this gets nasty very quickly, and as you astutely noticed, the first three coefficients of the Taylor Series expansion are $0$! It's not easy to readily notice a pattern from computing derivatives, and you might be wondering if the derivatives at $0$ evaluate to $0$ in this fashion forever, or if there are eventually non-zero coefficients in this Taylor expansion. Is there an easy way to see this, and an easy way to do this problem?
Well, yes, assuming you are familiar with the Taylor series for $\frac{1}{1-x}$.
Note that we can rewrite your expression as $\frac{1}{4}x^{3} \cdot \frac{1}{1-(\frac{1}{4}x^{2})}$. 
Recall that the Taylor series expansion of $\frac{1}{1-x}$ about $c = 0$ is given by 
$$\sum_{i=0}^{\infty} x^{i}$$
Hence, our Taylor Series expansion about $c = 0$ for the expression we're given is in fact: 
$$\frac{1}{4}x^{3} \cdot \sum_{i=0}^{\infty} (\frac{1}{4}x^{2})^{i}$$
You can easily see now why the coefficients on $1, x,$ and $x^{2}$ are $0$.
