Derivatives of a trace with respect to perturbation Let $\mathbf{X}$ and $\mathbf{Y}$ be symmetric positive-definite $n\times n$ matrices. Let $\{\lambda_i(\mathbf{A})\}$ denote the eigenvalues of $n\times n$ matrix $\mathbf{A}$ and $\mathbf{\Lambda}_{\mathbf{A}}$ denote the corresponding diagonalizing unitary matrix (if it exists) such that $\mathbf{\Lambda}_{\mathbf{A}}^T\mathbf{A}\mathbf{\Lambda}_{\mathbf{A}}=\mathbf{D}_{\mathbf{A}}$ where diagonal matrix $\mathbf{D}_{\mathbf{A}}$ contains $\{\lambda_i(\mathbf{A})\}$.
Let $\mathbf{Z}=\mathbf{X}+a\mathbf{Y}$ where $a>0$ is a scalar.  Clearly, $\mathbf{Z}$ is also symmetric positive-definite. By the usual definition of the application of functions to matrices, $\mathbf{Z}\log \mathbf{Z}=\mathbf{Z}\mathbf{\Lambda}_{\mathbf{Z}}\log(\mathbf{D}_{\mathbf{Z}})\mathbf{\Lambda}_{\mathbf{Z}}^T$, where $\log(\mathbf{D}_{\mathbf{Z}})$ contains logs of diagonal elements. Consequently, trace 
$$T=\operatorname{Tr}[\mathbf{Z}\log \mathbf{Z}]=\sum_{i=1}^n\lambda_i(\mathbf{Z})\log \lambda_i(\mathbf{Z}).$$
I am interested in the derivative of the trace with respect to $a$.  Is it possible to write expressions for $\frac{\partial T}{\partial a}$, $\frac{\partial^2 T}{\partial a^2}$, and $\frac{\partial^3 T}{\partial a^3}$ in terms of $a$ and eigenvalues $\{\lambda_i(\mathbf{X})\}$ and  $\{\lambda_i(\mathbf{Y})\}$ of $\mathbf{X}$ and $\mathbf{Y}$ (as well as $\mathbf{\Lambda}_{\mathbf{X}}$ and $\mathbf{\Lambda}_{\mathbf{Y}}$)?
 A: Here is a partial anwser.
In Section 2.5 of the Matrix Cookbook, there is an unnumbered formula between Eqn 98 and Eqn 99 which yields the derivative for the trace of $any$ matrix function.  
A few iterations with that formula should convince you that the $k$-th derivative of your function as
$$
(-1)^k \,{\rm tr}\big(Y^k Z^{1-k}\big)
$$
for $k>1$.  
For $k=1,\,$ the result includes a $\log$-term
$$
{\rm tr}\big(Y\log(Z)+Y\big)
$$
The complete answer involves reformulating these results in terms of the eigenvalues.
A: The answer to your question is "yes," but I guess you wanted to see what the expressions would be!
The Matrix Cookbook is your friend for this type of question. http://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf
You can use the equations in Section 2.3.
First, by the product rule:
$$ \frac{\partial T}{\partial a} = \sum_{i=1}^n \left[ \frac{\partial \lambda_i(\mathbf{Z})}{\partial a} \log \lambda_i(\mathbf{Z}) + \lambda_i(\mathbf{Z}) \frac{\partial \log \lambda_i(\mathbf{Z})}{\partial a} \right] $$
Next note that
$$ \frac{\partial \log \lambda_i(\mathbf{Z})}{\partial a} =
\frac{1}{\lambda_i(\mathbf{Z})} \frac{\partial \lambda_i(\mathbf{Z})}{\partial a} $$
so the only "tricky" thing is to compute the partial of the eigenvalues with respect to $a$.  That's basically Eqn. 67 in the Matrix Cookbook:
$$ \frac{\partial \lambda_i(\mathbf{Z})}{\partial a} = \mathbf{v_i}^T \frac{\partial \mathbf{Z}}{\partial a} \mathbf{v}_i $$
where $\mathbf{v}_i$ are the eigenvectors of $\mathbf{Z}$, i.e. are columns of $\mathbf{\Lambda}_\mathbf{Z}$.  Finally
$$ \frac{\partial \mathbf{Z}}{\partial a} = \mathbf{Y} $$.
You are on your own to combine the terms and to compute the higher derivatives from here. (An exercise for the reader.)
A: I know this answer comes years late, but I wanted to provide a thorough and correct response that the other answers are lacking. I'll point to Chapter 3 of Petz' book on matrix analysis as a reference.

Let $I\subseteq\mathbb{R}$ be an open interval and let $f:I\rightarrow\mathbb{R}$ be any differentiable function. We will make use of the divided differences of $f$, which are defined as follows. For two values $x,y\in I$, we define the first order divided differences as
$$
f^{[1]}(x,y) = \left\{\begin{array}{ll}
\displaystyle\frac{f(x)-f(y)}{x-y} & x\neq y\\
f'(x)& x=y.
\end{array}\right.
$$
We may extend the function $f$ to symmetric matrices whose eigenvalues are contained in $I$ by means of the spectral decomposition. That is, if $A$ is a matrix with spectral decomposition
$$
A = \sum_{i=1}^n \lambda_i\, v_iv_i^T
$$
with eigenvectors $V_1,\dots,v_n$ and eigenvalues $\lambda_1,\dots,\lambda_n\in I$, then we define $f(A)$ as
$$
f(A) = \sum_{i=1}^n f(\lambda_i)\, v_iv_i^T.
$$ 
We may assume without loss of generality that $A$ is diagonal, $A = \sum_{i=1}^n \lambda_i\, e_ie_i^T$, where $e_1,\dots,e_n$ are the standard basis vectors. Define the $n\times n$ matrix $D$ as the matrix of divided differences of $A$, whose $ij$th entry is defined as 
$$
D_{ij} = f^{[1]}(\lambda_i,\lambda_j).
$$ 
Let $B$ be any other $n\times n$ symmetric matrix. For all values of $t$ small enough, the eigenvalues of $A+tB$ will also be contained in $I$. Then
$$
\left.\frac{d}{dt}f(A+tB)\right|_{t=0} = D\circ B
$$
where $D\circ B$ is the entrywise product of $D$ and $B$ with matrix elements $(D\circ B)_{ij} = D_{ij}B_{ij}$. In particular, we find that
$$
\left.\frac{d}{dt}\operatorname{Tr}(f(A+tB))\right|_{t=0} = \operatorname{Tr}(D\circ B) = \sum_{i=1}^n D_{ii}B_{ii} = \sum_{i=1}^n f'(\lambda_i)B_{ii}.
$$
Interestingly, this depends only on the diagonal elements of $B$ (where we have assumed that $A$ is diagonal).
Higher-order derivtaives of this function may be taken assuming that the higher-order derivatives of $f$ exist. See for example Theorem 3.25 here (or section X.4 of Bhatia's book on Matrix Analysis).

In your question, you have chosen the function $f:(0,\infty)\rightarrow\mathbb{R}$ defined as $f(x)=x\log x$ for all $x\in(0,\infty)$. In this case, $f'(x)=1+\log(x)$ and 
$$
\left.\frac{d}{dt}\operatorname{Tr}(f(A+tB))\right|_{t=0} = \operatorname{Tr}(D\circ B) = \sum_{i=1}^n D_{ii}B_{ii} = \sum_{i=1}^n (1+\log\lambda_i)B_{ii}.
$$
