10. Suppose $f$ and $g$ are differentiable functions such that
$$
x g(f(x)) f^{\prime}(g(x)) g^{\prime}(x)=f(g(x)) g^{\prime}(f(x)) f^{\prime}(x)
$$
for all real $x$. Moreover, $f$ is nonnegative and $g$ is positive. Furthermore,
$$
\int_{0}^{a} f(g(x)) d x=1-\frac{e^{-2 a}}{2}
$$
for all reals $a$. Given that $g(f(0))=1$, compute the value of $g(f(4))$.
Answer: $e^{-16}$ or $\frac{1}{e^{16}}$
Solution: Differentiating the given integral with respect to $a$ gives $f(g(a))=e^{-2 a}$.
Now
$$
x \frac{d[\ln (f(g(x)))]}{d x}=x \frac{f^{\prime}(g(x)) g^{\prime}(x)}{f(g(x))}=\frac{g^{\prime}(f(x)) f^{\prime}(x)}{g(f(x))}=\frac{d[\ln (g(f(x)))]}{d x}
$$
where the second equals sign follows from the given. Since $\ln (f(g(x)))=-2 x$, we have $-x^{2}+C=\ln (g(f(x)))$, so $g(f(x))=K e^{-x^{2}}$. It follows that $K=1$ and $g(f(4))=e^{-16}$.