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?
 A: A nice way to do this is to use a somewhat stronger fact than the other answers: as a function of two variables, the expression $a^b$ is differentiable. Simply put, what this means is that, if $f(x)$ and $g(x)$ are functions, and $h(x)=f(x)^{g(x)}$, then $h'(x)$ is the sum of how fast this expression changes when we treat $f(x)$ as a constant plus how fast it changes when we treat $g(x)$ as a constant. You already know how to differentiate polynomials and exponentials, so this suffices - just apply both rules and add them!
Since we know that the derivative of $f^{g(x)}$ is $\log(f)\cdot f^{g(x)}\cdot g'(x)$ and the derivative of $f(x)^g$ is $g\cdot f(x)^{g-1}\cdot f'(x)$, we get
$$h'(x)=\underbrace{\log(f(x))\cdot f(x)^{g(x)}\cdot g'(x)}_{\text{Derivative treating $f$ as constant}}+\underbrace{g(x)\cdot f(x)^{g(x)-1}\cdot f'(x)}_{\text{Derivative treating $g$ as constant}}.$$
You would get the same result by using logarithmic differentiation as other answers suggest*, but I generally find this is a bit easier to remember and more generalizable - for instance, note that differentiating a product $f(x)\cdot g(x)$ can be done by this same method.

(*Logarithmic differentiation is a good way to prove this result, since you write $$\log(h(x)) = \log(f(x))\cdot g(x)$$then differentiate both sides to get $$\frac{h'(x)}{h(x)}=\frac{f'(x)\cdot g(x)}{f(x)}+\log(f(x))\cdot g'(x)$$ and moving $h(x)$ to the other side and substituting it for its formula gives the formula I claim)
A: 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*}
A: Since $(x+1)^{2x}=e^{2x\ln(x+1)}=e^{u(x)}$, its derivate is $u'(x)e^{u(x)}$. Notice that if $a:I\longrightarrow\mathbb{R}^{+*}$ and $b:I\longrightarrow\mathbb{R}$, then we define $a(x)^{b(x)}$ as $e^{b(x)\ln a(x)}$ for all $x\in I$
A: That whole $a^u$ thing works when $a$ is a constant, not another expression in terms of $x$.
To take the derivative of this, you would have to convert it to $\displaystyle e^{2x\ln(x+1)}$  and THEN use the Chain Rule. 
This would be $\displaystyle e^{2x\ln(x+1)}\left(2\ln(x+1)+\frac{2x}{x+1}\right)=(x+1)^{2x}\left(2\ln(x+1)+\frac{2x}{x+1}\right)$
A: Start with $$f(x) = a(x)^{b(x)}$$.
Now, we can approach this a few ways.  The one way is to go into multidimensional derivatives.  Define
$$h(x,y) = a(x)^{b(y)}$$
$$h'(x,y) = \left[ {\begin{array}{cc} 
\frac{ \partial h_x }{ \partial x } & \frac{ \partial h_x }{ \partial y } \\
\frac{ \partial h_y }{ \partial x } & \frac{ \partial h_y }{ \partial y } \\
\end{array} } \right] = \left[ {\begin{array}{cc} 
a'(x) b(y) a(x)^{b(y)-1} & 0 \\
0 & ln(a(x)) b'(y) a(x)^{b(y)}\\
\end{array} } \right]$$
Let $g(x) = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$.  Then $f(x) = h(g(x))$.  By the chain rule, $$f'(x) = h'(g(x)) g'(x) = h'(x,x) \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$$
or $$f'(x) = 
a'(x) b(x) a(x)^{b(x)-1} + ln(a(x)) b'(x) a(x)^{b(x)}$$
$$ f'(x) = f(x) * ( b(x) \frac{a'(x)}{a(x)} + b'(x) ln(a(x)))$$
We can then do a sanity check using logarithmic derivatives.  Also we can look at what happens when $a(x) = k$:
$$ f'(x) = f(x) * ( b(x) \frac{0}{a(x)} + b'(x) ln(k))$$
$$ f'(x) = ( b'(x) ln(k) ) k^{b(x)} $$
or $b(x)=k$:
$$ f'(x) = f(x) * ( k \frac{a'(x)}{a(x)} + 0 ln(a(x)))$$
$$ f'(x) = k a'(x) a(x)^{k-1}$$
which are both right.
Conceptually, this has two terms.  You add them to find out what the instant "percent change" in $f(x)$ will be.
$b \frac{a'}{a}$ is $b$ times the instant percent change in $a$:  If $a$ is growing by $1\%$, and you raise this to the power $b$, you grow by $b\%$.
$b' ln a$ is the change in $b$ wrt $x$ scaled by the log-scale of $a$.  With exponentials, a given linear-scale unit of exponent increase causes a scale by a factor in the base.  $ln(a)$ is the ratio between the two (the linear effect in the exponent to the exponential effect in the result), and we multiply that by $f$ because we are scaling the entire value.
