how to find the derivate of a function g(x) It's $g(x)={{x^{2}-1}\over{x^{2}+2}}$ and i have to calculate $g^{13}(0)$.
I can't calculate all the derivates so i think to use power series.
$g(x)={{x^2\over{x^{2}+2}}-{1\over{x^2+2}}}$
Can i use the geometric series?
 A: First perform a simple arithmetic trick
$$g(x) = \frac{x^2-1}{x^2+2}=1-\frac{3}{x^2+2}$$
then, since $g'(x) = \frac{6x}{(2 + x^2)^2}$ is a odd function, then odd-order derivatives must be odd functions so they are like
$$x \cdot p(x)$$ for a certain polynomial function such that $\deg[p(x)]$ is even.
Then conclude and find $g^{13}(0) = 0$ since $0$ is a root of the polynomial-rational odd-degree derivative.
A: In principle yes, but it would be easier if you were to perform a polynomial division
$$
\frac{x^2-1}{x^2+2}=1-\frac{3}{x^2+2}.
$$
Perhaps you can even divine from the symmetry of the function what odd derivatives are likely to be?
A: $$g(x)={{x^{2}-1}\over{x^{2}+2}}=\frac{(x^2+2)-2-1}{x^2+2}=1-\frac{3}{x^2+2}$$
$$=1-\frac{3}{2}(\frac{1}{1+(\frac{x}{\sqrt{2}})^2})$$
depending on the geometric series 
$$=1-\frac{3}{2}(1-(\frac{x}{\sqrt{2}})^2+(\frac{x}{\sqrt{2}})^4-(\frac{x}{\sqrt{2}})^6+.........$$
so every odd derivative at $x=0$ will be $0$ because all remaining terms will contain $x$ with power one or more
