5. [4] Let $f(x)=\sin ^{6}\left(\frac{x}{4}\right)+\cos ^{6}\left(\frac{x}{4}\right)$ for all real numbers $x$. Determine $f^{(2008)}(0)$ (i.e., $f$ differentiated 2008 times and then evaluated at $x=0$ ).
Answer: $\frac{3}{8}$ We have
$$
\begin{aligned}
\sin ^{6} x+\cos ^{6} x & =\left(\sin ^{2} x+\cos ^{2} x\right)^{3}-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right) \\
& =1-3 \sin ^{2} x \cos ^{2} x=1-\frac{3}{4} \sin ^{2} 2 x=1-\frac{3}{4}\left(\frac{1-\cos 4 x}{2}\right) \\
& =\frac{5}{8}+\frac{3}{8} \cos 4 x .
\end{aligned}
$$

It follows that $f(x)=\frac{5}{8}+\frac{3}{8} \cos x$. Thus $f^{(2008)}(x)=\frac{3}{8} \cos x$. Evaluating at $x=0$ gives $\frac{3}{8}$.