Given a square matrix $M$, we know the exponential of $M$ is $$\exp(M)=\sum_{n=0}^\infty{\frac{M^n}{n!}}$$
and the logarithm is $$\log(M)=-\sum_{k=1}^\infty\frac{(I-M)^k}{k}$$
The derivative of $\exp(M)$ should be itself. It is easy to prove if $\frac{dM}{M}=I$.
But how to calculate the derivative of $\log(M)$? By the same way of calculation of the derivative of $\exp(M)$, the derivative of $\log(M)$ cannot converge.
So what is the derivative of $\log(M)$?
The derivative of $\log(x)$ is $1/x$. The derivative of the power series $$ \sum_{n=1}^\infty (1-x)^n/n$$ is $$\sum_{n=1}^\infty (1-x)^{n-1}$$ which converges to $x^{-1}$ if $|x| < 1$. The matrix power series $$\sum_{n=1}^\infty (I-M)^{n-1}$$ converges to $M^{-1}$ if the spectral radius of $I-M$ is less than $1$.
However, caution is needed with the notion of "derivative" of a matrix valued function. The basic problem is that matrices don't always commute. You can't say that the derivative of $f(M)$ will be, e.g. $$ \lim_{A \to 0} (f(M+A) - f(M)) A^{-1}$$ For example, if $f(M) = M^2$, $$(f(M+A) - f(M)) A^{-1} = M + A M A^{-1} + A$$ and as $A \to 0$, $A M A^{-1}$ need not converge to anything in particular; it can do all sorts of strange things.