Answer
$$\frac{2 u 2^{u}}{\sqrt{u^{2}+1}} \left( \frac{1}{u}+ \ln 2 - \frac{u}{u^2+1}\right)$$
Work Step by Step
Given $$
y=\frac{2 u 2^{u}}{\sqrt{u^{2}+1}}
$$
Take $\ln $ for both sides
\begin{align*}
\ln y&= \ln \frac{2 u 2^{u}}{\sqrt{u^{2}+1}}\\
&=\ln 2 u 2^{u}- \ln \sqrt{u^{2}+1}\\
&=\ln 2u +\ln 2^u -\frac{1}{2}\ln (u^2+1)\\
&= \frac{1}{u}+ \ln 2 -\frac{1}{2} \frac{2u}{u^2+1}
\end{align*}
Then
\begin{align*}
\frac{y'}{y} &= \frac{1}{u}+ \ln 2 -\frac{1}{2} \frac{2u}{u^2+1}\\
&= \frac{1}{u}+ \ln 2 - \frac{u}{u^2+1}\\
y'&= y\left( \frac{1}{u}+ \ln 2 - \frac{u}{u^2+1}\right)\\
&=\frac{2 u 2^{u}}{\sqrt{u^{2}+1}} \left( \frac{1}{u}+ \ln 2 - \frac{u}{u^2+1}\right)
\end{align*}
