In my reference Page 520, Entropy and information, Quantum Computation and Quantum Information by Nielsen and Chuang, it is given that
The relative entropy $S(ρ||σ)$ is jointly convex in its arguments, where $S(ρ||σ)=tr(\rho\log\rho)-tr(\rho\log\sigma)$ and $\rho,\sigma$ are positive semidefinite with trace $1$.
Proof:
For arbitrary matrices $A$ and $X$ acting on the same space define $$ I_t(A,X)\equiv tr(X^\dagger A^tXA^{1-t})-tr(X^\dagger XA)\tag{11.98} $$ The first term in this expression is concave in $A$, by Lieb’s theorem, and the second term is linear in $A$. Thus, $I_t(A, X)$ is concave in $A$. Define $$ \color{red}{ \begin{align} I(A,X)&=\frac{d}{dt}\bigg|_{t=0}I_t(A,X)\\ &=tr(X^\dagger (\log A)XA)-tr(X^\dagger X(\log A)A) \end{align}}\tag{11.99} $$ Noting that $I_0(A,X)=0$ and using the concavity of $I_t(A,X)$ in $A$ we have \begin{align} I(\lambda A_1+(1-\lambda)A_2,X)&=\lim_{\Delta\to 0}\frac{I_\Delta(\lambda A_1+(1-\lambda)A_2,X)}{\Delta}\tag{11.100}\\ &\ge \lambda\lim_{\Delta\to 0}\frac{I_\Delta(A_1,X)}{\Delta}+\lim_{\Delta\to 0}\frac{I_\Delta(A_2,X)}{\Delta}\tag{11.101}\\ &=\lambda I(A_1,X)+(1-\lambda)I(A_2,X)\tag{11.102} \end{align} That is, $I(A,X)$ is a concave function of $A$. Defining the block matrices \begin{align} A\equiv\begin{bmatrix}\rho&0\\0&\sigma\end{bmatrix}\quad,\quad X\equiv\begin{bmatrix}0&0\\I&0\end{bmatrix}\tag{11.103} \end{align} we can easily verify that $I(A,X)=−S(ρ||σ)$. The joint convexity of $S(ρ||σ)$ follows from the concavity of $I(A,X)$ in $A$.
How do we reach $I(A,X)=\frac{d}{dt}\bigg|_{t=0}I_t(A,X)=tr(X^\dagger (\log A)XA)-tr(X^\dagger X(\log A)A)$ ?, where $I_t(A,X)\equiv tr(X^\dagger A^tXA^{1-t})-tr(X^\dagger XA)$
I think I am confused by the fact that we are taking the derivative of $A^t$ with respect to $t$ inside trace operation.
It looks like $\frac{d}{dt}A^t=\log(A)A^t$, why is that the case ?
It'd be helpful if someone could direct me in the right direction in order to understand the whole proof.