<s>[INST] Use a known Maclaurin series to derive a Maclaurin series for the indicated function. $$f(x)=x\cos(x)$$
I'm not quite sure how to do this. I did two others, which I presume is the right way to do it, as follows:
\begin{align}
e^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\\
e^{-x/4}&=1-\frac{x}{4}+\frac{x^2}{4^22!}-\frac{x^3}{4^33!}+\cdots\\
&=\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{4^nn!}
\end{align}
and
\begin{align}
\sin(x)&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\\
\sin(x^6)&=x^6-\frac{(x^6)^3}{3!}+\frac{(x^6)^5}{5!}-\frac{(x^6)^7}{7!}+\cdots\\
&=x^6-\frac{x^{18}}{3!}+\frac{x^{30}}{5!}-\frac{x^{42}}{7!}+\cdots\\
&=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{12n-6}}{(2n-1)!}
\end{align}
But where this one appears to be two separate functions multiplied together, I'm not sure how to approach it.
 [/INST] The well known Maclaurin series is $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \dots$
We can use this as follows, (look over @AWertheim's answer before viewing spoiler)

 $$ f(x) = x \cos x = x \left( 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \dots \right) = x - \dfrac{x^3}{2!} + \dfrac{x^5}{4!} - \dfrac{x^7}{6!} + \dots $$

</s>