What is the derivative of a function of the form $u(x)^{v(x)}$? So I have a given lets say $(x+1)^{2x}$ in addition to $\frac{\mathrm dy}{\mathrm dx}a^u=a^u\log(a)u'$. I still have to multiply this by the derivative of the inside function $x+1$ correct?

This is what logarithmic differentiation is for. You start with writing the function as an equation
$$y = (x + 1)^{2x},$$
then take the natural log of both sides:
$$\ln y = \ln\left[(x + 1)^{2x}\right] = 2x \ln(x+1).$$
We then implicitly differentiate both sides with respect to $x$. By chain rule (remember, $y$ is a function of $x$), the left side comes to
$$\frac{1}{y} \cdot y'.$$
The right side can be differentiated as normal:
$$\frac{2x}{x + 1} + 2\ln(x + 1).$$
So,
\begin{align*}
&\frac{1}{y} \cdot y' = \frac{2x}{x + 1} + 2\ln(x + 1) \\
\implies \, &y' = y\left(\frac{2x}{x + 1} + 2\ln(x + 1)\right) \\
\implies \, &y' = (x + 1)^{2x}\left(\frac{2x}{x + 1} + 2\ln(x + 1)\right).
\end{align*}
