The problem that I am looking at is the following perturbation problem from the notes on Trace Inequalities and Quantum Entropy on page 12, the following result is said to follow "by the Spectral Theorem and first order perturbation theory":
Let $f$ be a continuously differentiable function on $(0,\infty)$ and $B,C$ self-adjoint (complex) $n \times n$ matrices. Then $$\frac{d}{dt}\left|_{t=0}\right.\operatorname{Tr}[f(B + tC)] = \operatorname{Tr}[f'(B)C].$$
If someone knows a reference where I can learn why this is true (preferably an article, I could find online or through a University library), then that would be great.
Note: I already received a solution to this using some more powerful perturbation theory $C^1$ unitarily diagonalizing, but I was wondering about less complicated things.
The result is easy to check if $f(x)$ was a polynomial. For a general continuously differentiable function $f(x)$ and a given $\epsilon>0$, choose a polynomial $p(x)$ such that $|f(x)-p(x)|+|f'(x)-p'(x)|<\epsilon$ on an interval $I$ containing all of the eigenvalues of $B$ and $C$. One has $$\dfrac{d}{dt}Tr(p(B+tC))(0)=Tr(p'(B)C),$$ and so for $$g(t)=|Tr(p(B+tC))-Tr(p(B))-Tr(p'(B)C)t|$$ we have $g(t)/t \rightarrow 0$ as $t\rightarrow 0$. First note that $$|Tr(p'(B)C)-Tr(f'(B)C)|=|Tr((p'(B)-f'(B))C)| $$ $$\leq \|p'(B)-f'(B)\| \|C\|<n \epsilon \|C\|.$$ Moreover, $$|Tr(p(B+tC))-Tr(f(B+tC))|=|Tr(p(B+tC)-f(B+tC))|<n \epsilon,$$ and similarly $$|Tr(p(B))-Tr(f(B))|<n \epsilon.$$ Putting all these inequalities together, we have
$$h(t)=|Tr(f(B+tC))-Tr(f(B))-Tr(f'(B)C)t|\leq g(t)+(2+\|C\|)n\epsilon,$$ By letting $\epsilon \rightarrow 0$, we have $0\leq h(t)\leq g(t)$ which implies the formula.