Why does $\frac{d}{d\theta} \left(\theta\prod\limits_{i=1}^nx_i\right) = \sum\limits_{i=1}^n\ln x_i$ Is this just the product rule? I have this in my notes but I didn't think anything of it and now I'm wondering how this happens?

Edit:
Im working with maximum likelihood estimation and in my notes I have that the likelihood funciton $=L(x;\theta)=\prod_{i=1}^nf(x;\theta)$ where $x$ is the variable and $\theta$ is the parameter of a probability distribution. To estimate I was told that we take the log of the likelihood function, i.e. $\ln(L)$, then take its derivative to estimate the parameter. The function I'm working with is $f(x;\theta)=(\theta +1)x^{\theta}$. So 
$$
  L(x;\theta)=\prod_{i=1}^n(\theta +1)x_i^{\theta}=(\theta+1)^n\prod_{i=1}^nx_i^{\theta}.
$$ 
Now 
$$
  \ln(L(x;\theta))=n*\ln(\theta+1)+\theta \ln\left(\prod_{i=1}^nx_i\right).
$$
Here's where I'm confused, I have in my notes that 
$$
  \frac{d(\ln L)}{d\theta}=\frac{n}{\theta+1}+\sum_{i=1}^n\ln(x_i).
$$ 
Why does the product of $x_i$ become the summation of $x_i$?
$$
\ln(a\cdot b) = \ln(a) + \ln(b)
$$
This is a fundamental property of logarithms.
so lets use $$\prod_{i=0}^n x_i = P_n$$
then we have
$$
\ln\left(\prod_{i=0}^n x_i\right) = \ln\left(x_n\prod_{i=0}^{n-1} x_i\right) = \ln(x_n) + \ln\left(\prod_{i=0}^{n-1} x_i\right) = \ln(x_n) + \ln(P_{n-1})
$$
using induction or just seeing a pattern you find that we have
$$
\ln\left(\prod_{i=0}^n x_i\right)  = \sum_{i=0}^n\ln x_i
$$
$$ \ln(L(x;\theta))=n*\ln(\theta+1)+\theta \ln\left(\prod_{i=1}^nx_i\right)$$

$$=n*\ln(\theta+1)+\theta \ln\left(x_1x_2...x_n\right)$$

$$\stackrel{(*)}{=}n*\ln(\theta+1)+\theta [\ln(x_1) + \ln(x_2) + ... + \ln(x_n)]$$

$$=n*\ln(\theta+1)+\theta \sum_{i=1}^n\ln(x_i)$$

$$ \to \frac{d(\ln L)}{d\theta}=\frac{n}{\theta+1}+\sum_{i=1}^n\ln(x_i) $$ 

$(*)$ For real numbers $x_1, x_2, ..., x_n &gt; 0$, we have 

$$\ln\left(x_1x_2...x_n\right)=[\ln(x_1) + \ln(x_2) + ... + \ln(x_n)]$$