Chain rule for a composition with a curve

Statement

Let $I \subseteq \mathbb{R}$ be an open interval, let $U \subseteq \mathbb{R}^n$ be open, let $\gamma \colon I \to U$, let $f \colon U \to \mathbb{R}^m$, and let $t_0 \in I$. Suppose that $\gamma$ is differentiable at $t_0$ and $f$ is differentiable at $\gamma(t_0)$. Then $f \circ \gamma$ is differentiable at $t_0$, and $$(f \circ \gamma)'(t_0) = Df(\gamma(t_0))\big(\gamma'(t_0)\big).$$

Remarks

For $m = 1$, this reads $(f \circ \gamma)'(t_0) = \nabla f(\gamma(t_0)) \cdot \gamma'(t_0) = \sum_{i=1}^n \dfrac{\partial f}{\partial x_i}(\gamma(t_0)) \, \gamma_i'(t_0)$, with the gradient $\nabla f$. It is the case $n = 1$ of the chain rule in several variables, since a linear map $\mathbb{R} \to \mathbb{R}^n$ is $s \mapsto s \gamma'(t_0)$. For $\gamma(t) = a + t v$ it gives the directional derivative $D_v f(a) = Df(a)(v)$.

Explanations

L1 human

Walk along a curve $\gamma(t)$ through a landscape with height $f$. The rate at which your height changes is the slope of the landscape in the direction you walk, times your speed. For $f(x, y) = x y$ and $\gamma(t) = (t, t^2)$, the rule gives $\nabla f \cdot \gamma' = t^2 \cdot 1 + t \cdot 2t = 3 t^2$, the derivative of $t^3$.