25th derivation of $\cos{x^3}$ I have to calculate 25th derivative of function $f(x)=\cos{x^3}$ in $0$, $f^{(25)} (0)$. In my college, we usually use Newton-Leibnitz rule. We usually derivate it couple of times and then get something like $f^{(4)} = f^{(2)}x^2 + f^{(0)}$. This is not from this task, i am just giving you example.
 A: Recall that if $f$ has a Taylor series expansion $f(x) = \sum\limits_{n=0}^{\infty}{a_nx^n}$ in a neighborhood of $0$, then
$$ f^{(n)}(0) = a_nn!. $$
So it suffices to find the Taylor series of $\cos(x^3)$ and find the coefficient of $x^{25}$ in this expansion.
Since $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}+ \dots $$
we have
$$ \cos(x^3) = 1 - \frac{(x^3)^2}{2!} + \frac{(x^3)^4}{4!} - \frac{(x^3)^6}{6!} + \dots. $$
What is the coefficient of $x^{25}$ in the above expansion?
A: There is an easy way to do this. Try proving first that:

If $f$ is even, then $f'$ is odd, and that if $f$ is odd, then $f'$ is even.

Since $f$ is even, its first derivative is odd and its second derivative is even, etc. What does this tell you about $f^{(25)}$? What value must it take at $0$?
A: There are at least two solutions:
1) As already mentioned by Joey Zou, you can use Taylor series expansion of $f(x)=\cos{x^3}$ around $x=0$ to find its derivatives. Since $$f(x)=\cos{x^3}=\sum_{k=0}^{\infty}{(-1)^k\frac{(x^3)^{2k}}{(2k)!}}$$
or $$f(x)=1-\frac{x^6}{2!}+\frac{x^{12}}{4!}-\frac{x^{18}}{6!}+\frac{x^{24}}{8!}-\frac{x^{30}}{10!}+\dots$$
You can see that the $x^{25}$ term's coefficient is zero, and thus $f^{(25)}(0)=0$.
2) The function $f(x)=\cos{x^3}$ is an even function, i.e., $f(-x)=f(x)$. You can easily show that the derivative of an even function is an odd function and vice versa. Therefore, the 25th derivative of $f(x)$ is an odd function, and hence it has to be zero at $x=0$.
