Compute the 10th derivative of
at x=0
f^10(0)=?
i think im supposed to use the Maclaurin series
You get the McLaurin series up to the term in $\displaystyle x^{10}$ then you can differentiate term by term ten times (which is easy as the terms with exponent of $\displaystyle x$ less than $\displaystyle 10$ will be eliminated by the time you have taken the 10-th derivative) and then set $\displaystyle x=0$ to evaluate (the terms with powers greater than $\displaystyle 10$ when $\displaystyle x$ is set to zero will be zero so there was no need to look at them).
CB
Substitute $\displaystyle x^2/4$ for $\displaystyle u$ to get:
$\displaystyle \arctan(x^2/4)=\frac{x^2}{4}-{{x^6}\over{3.4^3}}+{{x^{10}}\over{5.4^5}}-{{x^{14}}\over{7.4^7}} + ..$
truncate after the term with $\displaystyle x^{10}$ as you won't need them (Why is that?) and diffterentiate 10 times (you do that), then substitute $\displaystyle 0$ for $\displaystyle x$ in what you have left (if anything).
CB
$\displaystyle \arctan(u) = u - \frac{u^{3}}{3} + \frac{u^{5}}{5} - \frac{u^{7}}{7} + ... $
$\displaystyle \arctan (x^{2}/4) = \frac{x^{2}}{4} - \frac{x^{6}}{3\cdot4^3} + \frac{x^{10}}{5 \cdot 4^{5}} + \frac{x^{14}}{7 \cdot4^{7}} + ... $
$\displaystyle \frac{d^{n}}{dx^{n}} x^{k} = \frac{k!}{(k-n)!} x^{k-n} $
The 10th derivatives of the first two terms are zero for any value of x, and the 10th derivative of every term beyond $\displaystyle x^{10} $ evaluates to zero at x=0.
so $\displaystyle \frac{d^{10}}{dx^{10}} \arctan \Big(\frac{x^{2}}{4}\Big) = \frac{1}{5 \cdot 4^5} \frac{d^{10}}{dx^{10}} x^{10} = \frac{1}{5 \cdot 4^{5}}\frac{10!}{0!} x^{0} = \frac{2835}{4}$
which of course at $\displaystyle x=0$ is $\displaystyle \ \frac{2835}{4} $