What is the derivative of $(u^v)$? The derivative of a sum is the sum of the derivatives, ie,
$$d(u+v)=du+dv$$
The derivative of a product is a little more complicated:
$$d(u\cdot v)= u\ dv + v\ du$$
But what about the derivative of a power? I'm not talking $x^n$ or $a^x$ or even $x^x$, but $u^v$ - an arbitrary function of $x$ raised to the power of another arbitrary function of $x$. 
 A: We start with
$$y=u^v$$
where $y$, $u$ and $v$ are all functions of $x$.
We take the natural logarithm $(\ln)$ of both sides: 
$$\ln(y)=\ln(u^v)$$
And drop $v$ out from $\ln$, using the relevant property of logarithms:
$$\ln(y)=v\ \ln(u)$$
This allows us to use the previously established product rule when we differentiate:
$$d\ln(y)=v\ d\ln(u) + \ln(u)\ dv$$
Which, by the calculus definition of the natural logarithm, simplifies to: 
$$\frac{dy}{y}=v\ \frac{du}{u} + \ln(u)\ dv$$
We then multiply both sides by $y$ to isolate $dy$, and replace $y$ with its definition $u^v$ to get:
$$dy=u^v\ \left(\frac{v\ du}{u} + \ln(u)\ dv\right)$$
Which is the answer we were looking for. But! Let's distribute that $u^v$ and do some cancelling:
$$dy=v\ u^{v-1}\ du + \ln(u)\ u^v\ dv$$
To the student of calculus, those look familiar. The first addend $(v\ u^{v-1}\ du)$ is the power rule $(d(u^n)=n\cdot u^{n-1}\ du)$, while the second $(\ln(u)\ u^v\ dv)$ is the general exponential rule $(dv=\ln(a)\ a^v\ dv)$.
This makes sense; the derivation of this rule used the product rule, which defines the derivative of a product as the sum of two distinct products, one each that differentiates one function and leaves the other alone.
Here, that's gone up a level, so to speak, with each addend of our result treating either the base or the power as constant. In fact, setting $u$ constant renders $du \; \; 0$, rendering the first addend $0$ in turn, and likewise for $v$ and the second addend, rendering both the power rule and the generalized exponent rule special cases of this "generalized power/exponent rule". 
A: A not so well known differentiation trick is to consider all variables in turn and assume the other to be constants, differentiate and sum.
Hence $u^v$ is first a power of $u$, then an exponential with exponent $v$ and
$$\left(u^v\right)'=vu^{v-1}u'+\log u\,u^vv'.$$

This is in fact an application of
$$\left(u^v\right)'=\frac{\partial u^v}{\partial u}u'+\frac{\partial u^v}{\partial v}v'.$$
A: Let $u=f(x)$, $v=g(x)$ and $y=f(x)^{g(x)}$. Then using the Chain and Product Rules, $$\begin{align}\ln y=g(x)\ln f(x)&\implies\frac1y\cdot\frac{dy}{dx}=g'(x)\ln f(x)+\frac{g(x)}{f(x)}\cdot f'(x)\\&\implies\frac{dy}{dx}=f(x)^{g(x)}\left(g'(x)\ln f(x)+\frac{g(x)}{f(x)}\cdot f'(x)\right)\end{align}$$ Hence $$d(u^v)=u^v\left(v'\ln u'+\frac{vu'}u\right)$$
A: 
We obtain
  \begin{align*}
\color{blue}{d(u^v)}&=d\left(e^{v\cdot \ln u}\right)\\
&=e^{v\cdot \ln u}\cdot d(v\cdot \ln u)\\
&=u^v\left(\ln u\ dv + v\ d\left(\ln u\right)\right)\\
&\,\,\color{blue}{=u^v\left(\ln u\ dv+\frac{v}{u}\ du\right)}
\end{align*}

