Chain Rule on a Curve¶

Theorem¶

Let $I\subseteq\mathbb{R}$ be an interval and let $\boldsymbol{\gamma}\colon I\to\mathbb{R}^n$ be a differentiable curve. Its velocity at $t_0$ is

\boldsymbol{\gamma}'(t_0)= \begin{pmatrix} \gamma_1'(t_0)\\ \vdots\\ \gamma_n'(t_0) \end{pmatrix}.

If $B\subseteq\mathbb{R}^n$ is an open set containing $\boldsymbol{\gamma}(I)$ and $f\colon B\to\mathbb{R}$ is differentiable at $\boldsymbol{\gamma}(t_0)\in B$ $B$ , then

\frac{\mathrm{d}}{\mathrm{d}t}(f\circ\boldsymbol{\gamma})(t_0) =\boldsymbol{D}f(\boldsymbol{\gamma}(t_0))\boldsymbol{\gamma}'(t_0) =\boldsymbol{\nabla} f(\boldsymbol{\gamma}(t_0))^{\top}\boldsymbol{\gamma}'(t_0).

Interpretation¶

A curve describes motion through a multivariable state space. The chain rule says the value changes by combining the system's velocity with the local sensitivity of the function of several variables.

A curve gamma(t) crossing circular level sets of f. At gamma(t0) the velocity vector and the gradient of f point in partly the same direction, so f increases along the curve there.

The chain rule gives $\frac{\mathrm{d}}{\mathrm{d}t}f(\boldsymbol{\gamma}(t))\big|_{t=t_0}=\boldsymbol{\nabla} f(\boldsymbol{\gamma}(t_0))\cdot\boldsymbol{\gamma}'(t_0)$ $γ^{'} (t_{0})$ . In the figure, the velocity has a positive component along the gradient, so $f$ increases along the curve at $t_0$ . The grey circles are level sets of $f$ .

A path assigns a position to each parameter value. Composing it with $f$ restricts $f$ to that motion; differentiating the composition pairs the derivative of $f$ with the path velocity.

Key Results¶

The derivative of $f(\boldsymbol{\gamma}(t))$ is the total derivative of $f$ applied to the velocity of $\boldsymbol{\gamma}$ .
If the velocity is a unit vector, the formula is a directional derivative.
The curve case is the one-dimensional input version of the multivariable chain rule.

Examples¶

If $f(x,y)=x^2+y^2$ and $\boldsymbol{\gamma}(t)=(t,2t)$ $(t, 2 t)$ , then $(f\circ\boldsymbol{\gamma})'(t)=10t$ $10 t$ .
If $f(x,y)=xy$ and $\boldsymbol{\gamma}(t)=(\mathrm{e}^t,t^2)$ $(e^{t}, t^{2})$ , then $(f\circ\boldsymbol{\gamma})'(t)=\mathrm{e}^t(t^2+2t)$ $2 t)$ .

Prerequisites¶

Sources¶

Math IV: Multivariate Calculus · Week 1, Lecture 2

Aliases: chain rule along a curve, curve chain rule, chain rule for curves, chain rule along a path Keywords: velocity along a curve, derivative along a path Symbols: element of, gamma, nabla gradient, real numbers, subset

Last updated: 22 October 2026 19:04