# Derivatives of component maps

Given functions $f_1:\mathbb{R}^{a_1}\rightarrow\mathbb{R}^{b_1}$ and $f_2:\mathbb{R}^{a_2}\rightarrow\mathbb{R}^{b_2}$, and the function $f:\mathbb{R}^{a_1+a_2}\rightarrow\mathbb{R}^{b_1+b_2}$ is defined by $$f(x,y)=(f_1(x),f_2(y))$$ for $x\in\mathbb{R}^{a_1},y\in\mathbb{R}^{a_2}$.

Take points $z=(z_1,z_2)$ and $w=(w_1,w_2)$, where $z_1,w_1\in\mathbb{R}^{a_1},z_2,w_2\in\mathbb{R}^{a_2}$.

Is it true that $Df(z)w=(Df_1(z_1)w_1, Df_2(z_2)w_2)$?

We have \begin{align*}\def\norm#1{\left\lVert#1\right\rVert} f(z+w) &= \bigl(f_1(z_1+w_1), f_2(z_2 + w_2)\bigr)\\ &= \bigl(f_1(z_1) + Df_1(z_1)w_1 + o(\norm{w_1}), f_2(z_2) + Df_2(z_2)w_2 + o(\norm{w_2}) \bigr)\\ &= f(z) + \bigl(Df_1(z_1)w_1, Df_2(z_2)w_2\bigr) + o(\norm{w}) \end{align*} As $w \mapsto \bigl(Df_1(z_1)w_1, Df_2(z_2)w_2\bigr)$ is linear in $w$, we have $Df(z)w = \bigl(Df_1(z_1)w_1, Df_2(z_2)w_2\bigr)$ by defnition.