14.108 min read

One Underlying Variable

An important special case: $\mathbf{u} = \mathbf{f}(t) \in \mathbb{R}^k$ depends on a single scalar variable $t$ , and $h = g(\mathbf{u}) = g(\mathbf{f}(t)) \in \mathbb{R}$ is a scalar output. This is the case $n=1$ , $m=1$ .

The Jacobian of $\mathbf{f}: \mathbb{R} \to \mathbb{R}^k$ is the column vector $\mathbf{f}'(t) = (f_1'(t), \ldots, f_k'(t))^T \in \mathbb{R}^{k\times 1}$ . The Jacobian of $g: \mathbb{R}^k \to \mathbb{R}$ is the row vector $\nabla^T g = (\partial g/\partial u_1, \ldots, \partial g/\partial u_k) \in \mathbb{R}^{1\times k}$ .

The chain rule gives $h'(t) = \nabla^T g(\mathbf{f}(t)) \cdot \mathbf{f}'(t) = \nabla g(\mathbf{f}(t)) \cdot \mathbf{f}'(t)$ — a dot product. This is a row vector times a column vector = a scalar.

Formal View

Theorem 14.6 — Chain Rule: One Underlying Variable

\mathbf{f}: \mathbb{R} \to \mathbb{R}^k

is differentiable at

t

and

g: \mathbb{R}^k \to \mathbb{R}

is differentiable at

\mathbf{f}(t)

, then

h(t) = g(\mathbf{f}(t))

satisfies

h'(t) = \nabla g(\mathbf{f}(t)) \cdot \mathbf{f}'(t) = \sum_{k=1}^K \frac{\partial g}{\partial u_k}(\mathbf{f}(t)) f_k'(t)

In Leibniz notation: $\frac{dh}{dt} = \sum_k \frac{\partial g}{\partial u_k}\frac{du_k}{dt}$ .

Why This Matters

The one-underlying-variable case is ubiquitous in physics where a function depends on space, and space depends on time.

Rate of change of a scalar quantity along a trajectory: $\frac{d}{dt}[f(\boldsymbol{\gamma}(t))] = \nabla f \cdot \boldsymbol{\gamma}'$
Hamiltonian mechanics: $\dot{H} = \nabla_\mathbf{q} H \cdot \dot{\mathbf{q}} + \nabla_\mathbf{p} H \cdot \dot{\mathbf{p}}$
Neural ODE: continuous-depth neural networks parameterized by a single "depth" variable

Learning Resources

Chain Rule: Scalar Output, One Parameter

Khan Academy

The chain rule when the underlying variable is a scalar.

9 min

Directional Derivative as a Chain Rule Application

MIT OpenCourseWare

Connecting the directional derivative formula to the chain rule.

40 min

Quiz

Question 1

If $h(t) = g(f_1(t), f_2(t))$ , then $h'(t)$ equals:

Common Mistakes

Forgetting to evaluate $\nabla g$ at $(f_1(t), f_2(t))$ , not at $(t, t)$ .
Treating $h'(t)$ as a vector when the output is scalar — it is a scalar.
Omitting the dot product: writing $\nabla g \cdot \mathbf{f}'$ as $\nabla g \mathbf{f}'$ without the dot, causing dimension confusion.