Derivative of a continuous linear map

Statement

Let $V$ and $W$ be normed vector spaces over $\mathbb{R}$ and let $A \colon V \to W$ be a bounded linear map. Then $A$ is Fréchet differentiable at each point $x \in V$, and its Fréchet derivative at $x$ is $A$ itself: $DA(x) = A$.

Remarks

The proof is one line: $A(x + h) - A(x) - A h = 0$ for each $h$, so the remainder is $0$. For $V = \mathbb{R}^n$ and $W = \mathbb{R}^m$, each linear map is bounded. So the derivative of $x \mapsto M x$, for an $m \times n$ matrix $M$, is the linear map with matrix $M$ at each point. An affine map $x \mapsto A x + b$ has the same derivative $A$ at each point, because the derivative of a constant map is $0$.

Explanations

L1 human

The best linear approximation of a linear map is the map itself. This is the analogue of the rule that the derivative of $x \mapsto c x$ is the constant $c$.