What is $\frac{\partial}{\partial{a}} (\prod_{i=1}^n x_i^a)$? Stuck on how to find this derivative:
$$\frac{\partial}{\partial{a}} (\prod_{i=1}^n x_i^a)$$
Any help?
 A: One may observe that
$$
\log\left( \prod_{i=1}^n x_i^a\right)=a \sum_{i=1}^n \log x_i
$$ giving, by differentiating with respect to $a$,
$$
\frac{\frac{\partial}{\partial{a}} (\prod_{i=1}^n x_i^a)}{(\prod_{i=1}^n x_i^a)}= \sum_{i=1}^n \log x_i
$$ and
$$
\frac{\partial}{\partial{a}} (\prod_{i=1}^n x_i^a)=\left(\sum_{i=1}^n \log x_i\right) \cdot (\prod_{i=1}^n x_i^a).
$$
A: $$\newcommand{\pder}{\frac{\partial}{\partial a}}$$
We want:
$$\pder\left(\prod_{i = 1}^n x_i^a\right)$$
The first thing to do is recall that $x_i^a = e^{a\ln(x_i)}$.  Rewrite them all this way to get that:
$$\pder\left(\prod_{i = 1}^n e^{a\ln(x_i)}\right)$$
Now, we're multiplying a bunch of exponentials together, so we add their exponents:
$$\pder\left(e^{\sum_{i = 1}^n a\ln(x_i)}\right)$$
We can factor out $a$, to get that:
$$\pder\left(e^{a\sum_{i = 1}^n\ln(x_i)}\right)$$
This term $\sum_{i = 1}^n \ln(x_i)$ can be rewritten as $\ln\left(\prod_{i = 1}^n x_i\right)$, but with respect to $a$ it's just a constant, so we might as well denote it $c$.
Then, we have that:
$$\pder\left(e^{ac}\right) = ce^{ac}$$
