I'm going over some basic statistical physics and I need to compute
$$\dfrac{\partial}{\partial \beta}\ln\sum_{i}e^{-\beta E_i}$$
Theres probably some simple trick i'm missing but i'm really struggling with this. My first thought was that the log would cancel the exponential and i'd be left with $\dfrac{\partial}{\partial \beta}\sum_{i}{-\beta E_i}$ so I can just differentiate term by term, but that isn't how logs work. Really I just need to figure out a way to take the log of a sum and I should be able to do the rest.
Any help would be appreciated, thanks.
Nothing can be done to "simplify" the logarithm of a sum.
However, we do not need to simplify in order to differentiate. Our function is $\ln\left(\sum e^{-\beta E_i} \right)$. So it has shape $\ln(g(\beta))$, where $g(\beta)=\sum e^{-\beta E_i}$. The partial derivative of $g(\beta)$ is not hard to compute. Thus by the Chain Rule our partial derivative is $$\frac{\sum -E_ie^{-\beta E_i}}{\sum e^{-\beta E_i}}.$$