Is the product rule true in a Banach algebra? Let $X$ be a Banach space and $\mathcal{L}(X)$ the Banach algebra of all bounded linear operators $L:X\to X$, where the norm is given by
$$\|L\|_\mathcal{L}=\sup\{\|L(x)\|_X;\;\|x\|_X=1\}$$
and the product is the composition of operators.
If $T,F:(0,\infty)\to\mathcal{L}(X)$ are differentiable functions, can we apply the product rule to derive $T(t)F(t)$?
Particularly, I'm interested in the case that $\{T(t)\}_{t\geq 0}$ and $\{F(t)\}_{t\geq 0}$ are both uniformly continuous semigroups of bounded linear operators.
Thanks.

Yes, you can. For $h \ne 0$,
$$
\begin{align}
    & \frac{1}{h}\left[T(t+h)F(t+h)-T(t)F(t)\right]\\
    & =\frac{1}{h}\{T(t+h)-T(t)\}F(t+h)+T(t)\frac{1}{h}\{F(t+h)-F(t)\}\\
    & =\frac{1}{h}\{T(t+h)-T(t)\}F(t) + T(t)\frac{1}{h}\{F(t+h)-F(t)\} \\
    & + \frac{1}{h}\{T(t+h)-T(t)\}\{F(t+h)-F(t)\}
\end{align}
$$
Therefore,
$$
\begin{align}
      & \left\|\frac{1}{h}\left[T(t+h)F(t+h)-T(t)F(t)\right]-\left[T'(t)F(t)+T(t)F'(t)\right]\right\| \\
     & \le \left\|\frac{1}{h}[T(t+h)-T(t)]-T'(t)\right\|\|F(t)\|
       +\|T(t)\|\left\|\frac{1}{h}[F(t+h)-F(t)]-F'(t)\right\| \\
     & + \left\|\frac{1}{h}[T(t+h)-T(t)]\right\|\left\|\frac{1}{h}[F(t+h)-F(t)]\right\||h|.
\end{align}
$$
The right side tends to 0 as $h\rightarrow 0$ assuming that $T$ and $F$ are differentiable at $t$.
