Prove $\frac{d}{dt} \Big|_{t=0}\mbox{tr}(e^{X+tY})=\mbox{tr}(e^XY)$ I’m asked to prove $\frac{d}{dt}\Big|_{t=0}\mbox{tr}(e^{X+tY})=\mbox{tr}(e^XY)$ for any $X,Y$ in $M_n(\mathbb{C})$. My attempt is to assume both $X$ and $Y$ are diagonalizable, and since the set of all diagonalizable matrices is dense in $M_n(\mathbb{C})$, if we can show this is true for diagonalizable matrices, then we are done. I expected this will somehow simplify the proof, but seems it does not work well unless I further assume $X,Y$ can be diagonalizable at the same time. Any suggestions on this?

For ease of typing, define
$$\eqalign{
A&= A(t) = X+tY
\quad\implies\quad
 &dA = Y\,dt \\
 &&X = \lim_{t\to 0}\,A(t) \\
}$$
The Frobenius product notation for the trace will also prove convenient
$$A:B = {\rm Tr}(A^TB) = B:A$$
First, note that the Taylor series is valid for the matrix exponential
$$\eqalign{
e^A = \sum_{k=0}^\infty \frac{A^k}{k!} 
\quad\implies\quad
de^A &= \sum_{k=0}^\infty\frac{1}{k!}\sum_{j=0}^{k-1} A^j\,dA\,A^{k-1-j} \\
}$$
Next, write the objective function and calculate its differential, gradient and derivative.
$$\eqalign{
\phi &= {\rm Tr}(Ie^A) \\
&= I:e^A \\
d\phi &= I:de^A \\
 &= I:\left[\sum_{k=0}^\infty\frac{1}{k!}\sum_{j=0}^{k-1} A^j\,dA\,A^{k-1-j}\right] \\
 &= \left[\sum_{k=0}^\infty\frac{1}{k!}\sum_{j=0}^{k-1} A^j\,I\,A^{k-1-j}\right]^T:dA \\
 &= \left[\sum_{k=0}^\infty\frac{1}{k!}\sum_{j=0}^{k-1} A^{k-1}\right]^T:dA \\
 &= \left[\sum_{k=0}^\infty\frac{A^k}{k!}\right]^T:dA
\qquad\implies \frac{\partial\phi}{\partial A} = \left[e^A\right]^T \\
 &= \left[e^A\right]^T:Y\,dt \\
\frac{d\phi}{dt} &= {\rm Tr}(e^AY) \\
}$$
Finally, take the limit as $t\to 0$.
$$\eqalign{
\lim_{t\to 0}\left(\frac{d\phi}{dt}\right)
 &= \lim_{t\to 0}\,{\rm Tr}\left(e^AY\right) \\ 
 &= {\rm Tr}\left(e^XY\right) \\
}$$

The rearrangement property of the Frobenius product was used in some of the steps above, e.g.
$$\eqalign{
A:BCD &= B^TA:CD \\
 &= AD^T:BC \\
 &= B^TAD^T:C \\
}$$
these follow directly from the cyclic property of the trace
$$
{\rm Tr}(ABCD)
 = {\rm Tr}(BCDA)
 = {\rm Tr}(CDAB) = etc.
$$

UPDATE
The above is an example of a formula (which can be found on page 12 of the Matrix Cookbook) which is valid for any analytic function $\;f(z)$
$$\eqalign{
\frac{\partial\,{\rm Tr}(f(A))}{\partial A} &= f^\prime(A)^T  \\
}$$
Let $f(z)=e^z$ then $f'(z)=e^z\;$ and the whole derivation reduces to
$$\eqalign{
\phi &= {\rm Tr}(e^A) \\
d\phi &= (e^A)^T:dA = {\rm Tr}(e^AY)\,dt \\
\frac{d\phi}{dt} &= {\rm Tr}(e^AY) \\
\lim_{t\to 0}\left(\frac{d\phi}{dt}\right) &= {\rm Tr}(e^XY) \\
}$$
