stuck on logarithm of derivative of sum $\frac{\partial\mathrm{log}(a+b)}{\partial a}$ I need to evaluate an expression similar to the following:
$\frac{\partial\mathrm{log}(a+b)}{\partial a}$
At this point I don't know how to proceed. $b$ is a constant so there should be some way to eliminate it.
How would you proceed in this case?
Actually, the original expression is much more complicated, and related to multiclass logistic regression, but I wanted to spare you tedious details.

Suppose you know $\frac{d}{dx} e^x=e^x$ then define $f(x)=e^x$ to have inverse function $g(x) = \ln (x)$ to be the inverse function of the exponential function. In particular this means we assume $e^{\ln(x)}=x$ for all $x \in (0, \infty)$ and $\ln(e^x) = x$ for all $x \in (-\infty, \infty)$. The existence of $g$ is no trouble as the exponential function is everywhere injective. 
Ok, assume $x>0$ and let $y=\ln(x)$ then $e^y = e^{\ln(x)}=x$. Differentiate the equation $e^y=x$:
$$ \frac{d}{dx} e^y = \frac{d}{dx} x \ \ \Rightarrow \ \ e^y \frac{dy}{dx}=1$$
where we have used the chain-rule and the observation that $y$ is a function of $x$. Finally, solve for $\frac{dy}{dx}$ (which is what we're after here)
$$ \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x} \ \ \Rightarrow \ \ \frac{d}{dx} \ln (x) = \frac{1}{x}.$$
Now, to solve your problem I merely apply this to $\ln(a+b)$ thinking of $a$ as $x$.
$$ \frac{\partial}{\partial a} \ln (a+b) = \left[\frac{d}{dx} \ln (x+b) \right]_{x=a} =
\left[\frac{1}{x+b}  \right]_{x=a} = \frac{1}{a+b}.$$ 
