Derivative of a map into a product

Statement

Let $V$, $W_1$ and $W_2$ be normed vector spaces over $\mathbb{R}$, let $U \subseteq V$ be open, and let $x \in U$. Let $f_1 \colon U \to W_1$ and $f_2 \colon U \to W_2$ be Fréchet differentiable at $x$. Then the map $F \colon U \to W_1 \times W_2$, $F(y) = (f_1(y), f_2(y))$, is Fréchet differentiable at $x$, and $$DF(x)(h) = \big(Df_1(x)(h), \, Df_2(x)(h)\big) \quad \text{for each } h \in V.$$ Here $W_1 \times W_2$ has the norm $\|(w_1, w_2)\| = \max(\|w_1\|, \|w_2\|)$.

Remarks

The converse also holds: if $F$ is differentiable at $x$, then so are its two components, because each projection $W_1 \times W_2 \to W_i$ is a bounded linear map. Any norm on $W_1 \times W_2$ that is equivalent to the max norm gives the same derivative. Applied $m$ times with $W_i = \mathbb{R}$, this gives the derivative of a map into R^n by components.

Explanations

L1 human

To differentiate a map with two outputs, differentiate each output and put the two results side by side. The remainder of the pair is small exactly when both remainders are small.