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?
 A: We may assume $X \neq 0$. Put $f_m(t)= \frac{1}{m!}\mbox{tr}(X+tY)^m$. Then $\mbox{tr}(e^{X+tY})=\sum_{m=0}^{\infty} f_m(t)$. We will interchange the differentiation and the infinite sum.
Let's find an upper bound for $|f_m'(t)|$ using the Frobenius norm $\lVert \cdot \rVert$. Recall that $|\mbox{tr}(A)| \leq \sqrt{n} \cdot \lVert A \rVert$ hold for any $A \in M_n(\mathbb C)$. Pick $a>0$ and suppose $t \in [-a, a]\setminus \{0\}$. Then
\begin{align*} |f_m'(t)| &\leq  \frac{\sqrt{n}}{m!}\sum_{r=1}^{m} \binom{m}{r}r|t|^{r-1} \lVert X\rVert^{m-r} \lVert Y\rVert^r  \\ &\leq \frac{\sqrt{n}}{m!}\sum_{r=1}^{m} \binom{m}{r}ra^{r-1} \lVert X\rVert^{m-r} \lVert Y\rVert^r \\ &=\sqrt{n}\cdot\frac{\lVert X \rVert^m}{m!}\sum_{r=1}^{m} \binom{m}{r}ra^{r-1} \lVert X\rVert^{-r} \lVert Y\rVert^r \\&= \sqrt{n}\cdot\frac{\lVert X \rVert^m}{m!} \frac{d}{da} (1+a\lVert X \rVert^{-1} \lVert Y \rVert )^m \\&= \frac{\sqrt{n} \cdot \lVert Y \rVert}{(\lVert X \rVert +a\lVert Y \rVert )}  \cdot\frac{1}{m!} (\lVert X \rVert+a \lVert Y \rVert )^{m}  \\ &:=M_m\end{align*}
Observe that \begin{align*}\sum_{m=0}^{\infty} M_m = \frac{\sqrt{n} \cdot \lVert Y \rVert}{(\lVert X \rVert +a\lVert Y \rVert )}  \cdot\exp(\lVert X \rVert+a \lVert Y \rVert) < \infty \end{align*}
By the Weierstrass M-test, $\sum_{m=0}^{\infty}f_m'$ converges uniformly in $[-a, a] \setminus \{ 0\}$. If $t=0$, then \begin{align} &f_m'(0)=\frac{\mbox{tr}(X^{m-1}Y)}{(m-1)!} \tag{if  $m \geq 1$} \\ &f_0'(0) = 0 \end{align}
whence $\sum_{m=0}^{\infty} f_m'(0)= \mbox{tr}\left(e^X Y \right)$. To sum up, $\sum_{m=0}^{\infty}f_m'$ converges uniformly on $[-a, a]$. It follows that $f(t):=\mbox{tr}(e^{X+tY})=\sum_{m=0}^{\infty}f_m(t)$ satisfies $f'(t) = \sum_{m=0}^{\infty} f_m'(t)$ for all $t \in [-a, a]$. In particular, $f'(0)=\sum_{m=0}^{\infty}f_m'(0) = \mbox{tr}(e^X Y)$.
A: The trace is just the sum of diagonal values, so you can put the derivative inside the trace:
$$\frac{d}{dt}|_{t=0}tr(e^{X+tY})=tr(\frac{d}{dt}|_{t=0}e^{X+tY})=tr(e^{X+tY}Y)|_{t=0}=tr(e^XY)$$
A: 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
$$\eqalign{
\phi &= {\rm Tr}(e^A) \\
d\phi &= (e^A)^T:dA = {\rm Tr}(e^AY)\,dt \\
\frac{d\phi}{dt} &= {\rm Tr}(e^AY) \\
\lim_{t\to 0}\left(\frac{d\phi}{dt}\right) &= {\rm Tr}(e^XY) \\
}$$
A: Truly, definitely, only a side-comment, irrelevant to anybody's immediate needs, but a comment worth making - I think! maybe? I hope??? - and one truly, definitely, too much of a pain to enter in the comment section.
From the Lie theoretic point of view,
$${d\over dt}\Bigg\vert_{t=0}\exp (X+tY)=  d\exp_X(Y),$$
where $d\exp_X$ denotes the derivative of the (Lie theoretic!) exponential function
$$\exp\colon {\mathfrak g} \to G,$$
at $X\in {\mathfrak g} = M_n({\mathbb C})$,
with $G = GL_n({\mathbb C})$, the invertible matrices,  and $Y\in T_X{\mathfrak g}$.
It turns out that, suitably interpreted,
$$ d\exp_X (Y)= g(\text{ad }X)Y \cdot \exp X, $$
where ad is the adjoint operator, and
$$g(z) = {e^z -1 \over z  }.$$
("Suitably interpreted": for instance, above, the multiplication on the right by $\exp X$ on the tangent space is the 'functorial' operation  arising from multiplication on the right on the Lie group.)
All of this turns up - for instance - in the (or 'a') proof of the Hausdorff-Campbell-Baker formula.  For a non-power series proof (sort of), albeit one arguably a bit heavy on notation (but certainly more explicit than mine), see 4.26, and the following few sections (this answer/comment is basically lemma 4.27) of  Natural Operations in Differential Geometry (Kolar, Michor, Slovak) 
Taking traces 'kills' ad, and one ends up with your formula, almost, except that $\exp X$ is on the right - but, that doesn't matter. (I did it that way in keeping with this reference.)
