Derivation Gaussian Mixture Models log-Likelihood I'm trying to understand the derivation of the log-likelihood function for Gaussian Mixture Models. According to my records the following steps are made.
The log-likelihood function is defined as:
$
L(X|\Theta) = \sum_{n=1}^N ln P(x_n|\Theta) = \sum_{n=1}^N ln \sum^M_{m=1} \alpha_m \mathcal{N}(x_n|\mu_m, \Sigma_m)
$
And then the partial derivative w.r.t. $\mu_m$ is made, where the first step seems fine for me, applying the chain rule and the derivative of the log:
$$
\frac{\partial ln P(X|\theta)}{\partial \mu_m} = \sum_{n=1}^N \frac{1}{\sum_{m'=1}^M\alpha_{m'} \mathcal{N}(x_n|\mu_{m'},\Sigma_{m'})} \cdot \frac{\partial \sum^M_{m=1} \alpha_m \mathcal{N}(x_n|\mu_m, \Sigma_m)}{\partial\mu_m}
$$
I just don't get how the prior equation leads to the following: 
$$
\frac{\partial ln P(X|\theta)}{\partial \mu_m} = 
\sum_{n=1}^N \frac{\alpha_m \mathcal{N}(x_n|\mu_n, \Sigma_m)}{\sum_{m'=1}^M\alpha_{m'} \mathcal{N}(x_n|\mu_{m'},\Sigma_{m'})} \cdot
\frac{\partial[ln(\alpha_m) + ln \mathcal{N}(x_n|\mu_m, \Sigma_m)]}{\partial\mu_m}
$$
From where does the log arises here again after the derivation?
 A: Continuing from where you left off we have:
\begin{equation*}
   \frac{\partial ln P(X|\theta)}{\partial \mu_m} = \sum_{n=1}^N \frac{1}{\sum_{m'=1}^M\alpha_{m'} \mathcal{N}(x_n|\mu_{m'},\Sigma_{m'})} \cdot \frac{\partial \,[\alpha_m \mathcal{N}(x_n|\mu_m, \Sigma_m)]}{\partial\mu_m}
\end{equation*}
Now we make use of the following "trick":
\begin{equation*}
\frac{\partial\,[f(x)]}{\partial x} = f(x)\cdot \frac{\partial\,[\ln\left(f(x)\right)]}{\partial x}
\end{equation*}
This leads to the result
\begin{align*}
   \frac{\partial ln P(X|\theta)}{\partial \mu_m} &= \sum_{n=1}^N \frac{1}{\sum_{m'=1}^M\alpha_{m'} \mathcal{N}(x_n|\mu_{m'},\Sigma_{m'})} \cdot \alpha_m \mathcal{N}(x_n|\mu_m, \Sigma_m) \cdot \frac{\partial \,[\ln\left(\alpha_m \mathcal{N}(x_n|\mu_m, \Sigma_m)\right)]}{\partial\mu_m} \\[18pt]
&= \sum_{n=1}^N \frac{\alpha_m \mathcal{N}(x_n|\mu_m, \Sigma_m)}{\sum_{m'=1}^M\alpha_{m'} \mathcal{N}(x_n|\mu_{m'},\Sigma_{m'})} \cdot  \frac{\partial \,[ln(\alpha_m) + ln( \mathcal{N}(x_n|\mu_m, \Sigma_m))]}{\partial\mu_m}
\end{align*}
This is how the log was reintroduced in the derivation process.
A: It is basically wrong. You can think the derivative of a gaussian with respect to mean as a derivative of 
$$ ae^{bx} $$
with respect to x
so it should be $$ ae^{bx}\cdot b$$
which actually corresponds to
$$\frac{\partial ln P(X|\theta)}{\partial \mu_m} = 
\sum_{n=1}^N \frac{\alpha_m \mathcal{N}(x_n|\mu_m, \Sigma_m)}{\sum_{m'=1}^M\alpha_{m'} \mathcal{N}(x_n|\mu_{m'},\Sigma_{m'})} \cdot \Sigma_{m}^{-1}(x_n-\mu_m)
$$
