Derivative of Product of Operator-Valued Functions

Theorem
Let $\struct {X, \norm \cdot_X}$, $\struct {Y, \norm \cdot_Y}$, and $\struct {Z, \norm \cdot_Z}$ normed vector spaces.

Let $B(X,Y)$, $B(Y,Z)$, and $B(X,Z)$ denote the space of bounded linear transformations between $X$ and $Y$, between $Y$ and $Z$, and between $X$ and $Z$, respectively.

Let $A : \R \to B(X,Y)$ and $B : \R \to B(Y,Z)$ differentiable functions with values in the bounded linear transformations.

The product $AB : \R \to B(X,Z)$, $x \mapsto A(x)B(x)$ is differentiable and its derivative at any $x\in\R$ is given by $A'(x)B(x)+A(x)B'(x)$.

Proof
Given any $x,h\in X$, $h\neq 0$ we compute $ \norm{\frac {\map {(AB)} {x + h} - \map {(AB)} x} h-\map {A'} x \map B x - \map A x \map {B'} x}_{B(X,Z)} $ to be

$\norm{\frac {\map {A} {x + h} - \map {A} x } h-\map {A'} x }_{B(X,Y)}$ vanishes as $h\to 0$ because $\map{A'}x$ is the derivative of $A$ at $x$ which exists by assumption.

$\norm{\frac {\map {B} {x + h} - \map {B} x } h-\map {B'} x }_{B(Y,Y)}$ vanishes as $h\to 0$ because $\map{B'}x$ is the derivative of $B$ at $x$ which exists by assumption.

$\norm{ \map {B} {x + h} - \map B x }_{B(Y,Z)}$ vanishes as $h\to 0$ because

$\norm{ \map {B} {x + h} }_{B(Y,Z)}\to \norm{ \map {B} x }_{B(Y,Z)}$ as $h\to 0$ by the Reverse Triangle Inequality

Together this shows
 * $\ds \lim_{h \mathop \to 0}\norm{\frac {\map {(AB)} {x + h} - \map {(AB)} x} h-\map {A'} x \map B x - \map A x \map {B'} x}_{B(X,Z)}=0$

Therefore $A'(x)B(x)+A(x)B'(x)$ is the derivative of $AB$ at $x$.