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?

The trace is just the sum of diagonal values, so you can put the derivative inside the trace:
$$\frac{d}{dt}|_{t=0}tr(e^{X+tY})=tr(\frac{d}{dt}|_{t=0}e^{X+tY})=tr(e^{X+tY}Y)|_{t=0}=tr(e^XY)$$
