14.38 min read

Univariate Chain Rule

The familiar single-variable chain rule says: if $h(x) = g(f(x))$ , then $h'(x) = g'(f(x)) \cdot f'(x)$ .

In terms of Jacobians: $Dh(a) = Dg(f(a)) \cdot Df(a)$ . Since $n=k=m=1$ , all Jacobians are $1\times 1$ matrices (scalars), and Jacobian multiplication is just scalar multiplication. The univariate chain rule is the $1\times 1$ matrix case of the general chain rule.

The key insight from this perspective: the intermediate variable $u = f(a)$ appears as the evaluation point for $g'$ . This is why we write $g'(f(x))$ , not $g'(x)$ — we must differentiate $g$ at the intermediate value $u = f(x)$ , not at $x$ itself.

Formal View

Theorem 14.1 (Univariate Chain Rule) — Chain Rule for Scalar Functions

f: \mathbb{R} \to \mathbb{R}

is differentiable at

a

and

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

is differentiable at

f(a)

, then

h = g \circ f

is differentiable at

a

and

h'(a) = g'(f(a)) \cdot f'(a)

In Leibniz notation: $\frac{dh}{dx} = \frac{dg}{du}\cdot\frac{du}{dx}$ where $u = f(x)$ . The "cancellation of $du$ " is a useful mnemonic, though not literally true.

Interactive Visualization

Chain Rule Circuit Diagram

Why This Matters

The univariate chain rule is the foundation for all differentiation of compositions — mastering it is essential.

Differentiating $e^{x^2}$ , $\sin(\cos x)$ , $\ln(1+x^2)$ and all composed scalar functions
Implicit differentiation: treating $y$ as a function of $x$ in $F(x,y)=0$
Solving ODEs by substitution: $u = f(x)$ transforms the equation

Learning Resources

Chain Rule

3Blue1Brown

Essence of Calculus episode on the chain rule with beautiful geometric intuition.

17 min

Chain Rule — Examples

Khan Academy

Multiple worked examples of the univariate chain rule.

12 min

Quiz

Question 1

If $h(x) = \sin(x^2)$ , then $h'(x)$ equals:

Question 2

In the chain rule $h'(x) = g'(f(x))\cdot f'(x)$ , where is $g'$ evaluated?

Common Mistakes

Evaluating $g'$ at $x$ instead of at $f(x)$ — the outer derivative must be evaluated at the intermediate variable.
Forgetting to multiply by $f'(x)$ — every chain rule application adds a factor.
Confusing the chain rule with the product rule for $f(x)g(x)$ .