Derivative of a smooth map between Euclidean spaces Let $f:U\subset \Bbb R^k \to \Bbb R^l$ be a $C^\infty$ map, where $U$ is open. For $x\in U$ and $h\in \Bbb R^k$ define $df_x:\Bbb R^k \to \Bbb R^l$ by $df_x(h)=\lim_{t\to 0} (f(x+th)-f(x))/t$. How can we show that this map is linear in $h$ and that it corresponds to the $l \times k$ matrix $((\partial f_i/\partial x_j)|_x)$?
Taking $h=e_j$, we obviously get $df_x(e_j)=(\partial f_1/\partial x_j|_x,...,\partial f_l/\partial x_j|_x)$, so it suffices to show that $df_x$ is linear in $h$, but I can't see why. Thanks in advance.
 A: Your $df_x(h)$ is known as the directional derivative of $f$ at $x$ in direction $h$. Let $Df_x$ denote the derivative of $f$ at $x$ which is the linear map $\mathbb R^k \to \mathbb R^l$ represented by the Jacobian matrix $\left(\frac{\partial f_i}{\partial x_j}  \mid_x \right)$ of $f$ at $x$. It is well-known that $Df_x(h) = df_x(h)$. See any textbook, e.g. Rudin's "Principles of Mathematical Analysis" p.217/218.
This shows that $df_x$ is linear.
Let us nevertheless give a proof.
For $x, h \in \mathbb R^k$ define
$$\phi : \mathbb R \to \mathbb R^k, \phi(t) = x + th .$$
This is a $C^\infty$-function such that $\phi(0) = x$ and $\phi'(0) = h$. There exists an open interval $J$ containing $0$ such that $\phi(J) \subset U$. Let $\psi : J \to U$ denote the restrictuon of $\phi$ and define
$$g = f \circ \psi : J \to \mathbb R^l .$$
This is a $C^\infty$-function such that $g(t) = f(x+th)$. We have
$$df_x(h) = g'(0) = Dg_0(1) .$$
To understand the last equation, note that the linear map $Dg_0: \mathbb R \to \mathbb R^l$ is represented by the Jacobian $\left(\frac{\partial g_i}{\partial t}  \mid_0 \right) = \left( g_i'(0) \right) = g'(0) \in \mathbb R^l$, thus $Dg_0(1) = g'(0)$.
Now the chain rule for multivariable functions gives us
$$Dg_0 = Df_{g(0)} \circ D\psi_0 = Df_x \circ D\psi_0$$
which shows
$$df_x(h) = Df_x(D\psi_0(1)) .$$
But similarly as for $g$ we see that
$$D\psi_0(1) = \psi'(0) = \phi'(0) = h .$$
