Derivative of Tr$(\rho\log\rho)$

Question

The von Neumann entropy is the analogue of the Shannon entropy and is defined for positive semidefinite matrices as

$$S = \text{Tr}(\rho\log\rho)$$

It is computed in terms of the eigenvalues $\lambda_i$ as $S = \sum_i\lambda_i\log\lambda_i$ where we set $0\log(0) = 0$.

1. Is $S$ differentiable with respect to $\rho$? The answer to this seems to be yes (see for example the question here and links therein) and is given as $S'(\rho) = I + \log\rho$.

2. If the result above is true, what happens when $\rho$ has at least one eigenvalue of zero? The von Neumann entropy is defined for positive semidefinite matrices while the log is only defined for positive definite matrices.