Christmas Streak 30/88: Sum of... wait, what?

This is a very quick and undetailed solution. If you have any doubts, ask in the comments section.

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Differentiate both sides of $F(g(x))=\dfrac{1}{2}F(x).$ Then $g '(x)f(g(x))=f(x).$

Let $p=(4n-3)\pi$ and $q=(4n-1)\pi.$

And observe that $y=f(x)$ is symmetrical over the lines $x=\dfrac{p}{2}$ and $x=\dfrac{q}{2}.$

Therefore $\displaystyle \int_{0}^{\frac{1}{2}p}f(t)dt=\frac{1}{2}\int_{0}^{p}f(t)dt$ and $\displaystyle \int_{0}^{\frac{1}{2}q}f(t)dt=\frac{1}{2}\int_{0}^{q}f(t)dt.$

Take a closer look at $F(g(x))=\dfrac{1}{2}F(x).$ Since $\displaystyle F(x)=\int_{0}^{x}f(t)dt,$ we can say that $\displaystyle \int_{0}^{g(x)}f(t)dt=\frac{1}{2}\int_{0}^{x}f(t)dt.$

Note that $f(x)$ is always positive and therefore $F(x)$ is strictly increasing over all reals.

So plugging in $x=p$ in the equation, we see that $g(p)=\dfrac{1}{2}p.$

$g '(p)f(g(p))=\dfrac{1}{2}f(p)~\Rightarrow~g '(p)=\dfrac{1}{3}.$

Doing the same with $q,$ we get $g '(q)=1.$

$\displaystyle \therefore~\sum_{n=1}^{30}g '((2n-1)\pi)=\sum_{n=1}^{15}\{g '((4n-3)\pi)+g '((4n-1)\pi)\}=15(g '(p)+g '(q))=\boxed{20}.$