Unit 3: Differentiation — Composite, Implicit & Inverse Functions

The Chain Rule

Differentiating composite functions using the chain rule.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Differentiating a Function Wrapped Inside Another

None of the rules so far can differentiate (3x + 1)⁵ directly — it isn't a sum, product, or quotient of simple power terms; it's one function, u⁵, wrapped around another, 3x + 1. Before reading on: if y changes m times as fast as some in-between quantity u, and u itself changes n times as fast as x, how fast should y change relative to x?

The answer is m times n — rates of change compound by multiplication when one function feeds into another. That's the intuition behind the chain rule: differentiating a composite function requires the outer function's derivative, evaluated at the inner function's value, multiplied by the inner function's own derivative.

Definition — The Chain Rule

For a composite function f(g(x)), the derivative is d/dx[f(g(x))] = f'(g(x)) · g'(x) — the outer function's derivative, evaluated at the inner function, times the inner function's derivative.

\frac{d}{dx}\big[f(g(x))\big] = f'(g(x)) \cdot g'(x)

Worked Example — Differentiating a Power of a Function

Differentiate y = (3x + 1)⁵. The outer function is f(u) = u⁵ with f'(u) = 5u⁴; the inner function is g(x) = 3x + 1 with g'(x) = 3. By the chain rule: dy/dx = 5(3x + 1)⁴ · 3 = 15(3x + 1)⁴.

Worked Example — Differentiating a Trig Function of a Function

Differentiate y = sin(x²). The outer function is f(u) = sin(u) with f'(u) = cos(u); the inner function is g(x) = x² with g'(x) = 2x. By the chain rule: dy/dx = cos(x²) · 2x = 2x cos(x²).

Worked Example — Verifying a Chain Rule Result Numerically

Differentiate y = √(x² + 1) = (x² + 1)^(1/2). Outer: f(u) = u^(1/2), f'(u) = (1/2)u^(−1/2). Inner: g(x) = x² + 1, g'(x) = 2x. dy/dx = (1/2)(x² + 1)^(−1/2) · 2x = x/√(x² + 1). Check numerically at x = 1: the formula gives 1/√2 ≈ 0.7071. Estimating directly, y(1.001) = √2.002001 ≈ 1.41493 and y(1) = √2 ≈ 1.41421, so [y(1.001) − y(1)]/0.001 ≈ 0.707 — matching.

Derivative Explorer

f(x) =

x^3 - 3*x

Point of tangency

x = a1

Show tangent line

Trace mode

f(x), with tangent line at x = a

Show f''(x)Show f'''(x)

Derivative graph — linked to the graph above; panning or zooming either moves both

What's happening at x = a

At x = 1, f(x) ≈ -2 — the point (1, -2).
The derivative there is f'(1) ≈ 0: the slope of the tangent line, and the instantaneous rate of change of f at this exact point.
That slope is positive, so f is increasing at this point.
The second derivative is positive here, so f is concave up (curving upward) near this point.
Near x = 1, the tangent line y ≈ -2 + 0·(x − 1) is f's best straight-line approximation — the core idea a derivative captures.

Numerical derivative comparison

Forward, backward, and central difference estimates of f'(a) as the step size h shrinks — all three should converge toward the same value.
h	Forward	Backward	Central
1	4	-2	1
0.1	0.31	-0.29	0.01
0.01	0.0301	-0.0299	0.0001
0.001	0.003	-0.003	0

Analysis of f(x)

y-intercept: (0, 0)
x-intercepts: (-1.73, 0), (0, 0), (1.73, 0)
Extrema: local max at (-1, 2); local min at (1, -2)
Inflection points: (0, 0)
Vertical asymptotes: none found in view
Horizontal asymptotes: none found
Domain: all real numbers in view
Range (estimated): approximately [-970, 970]

Analysis of f'(x) — its roots are f's critical points

y-intercept: (0, -3)
x-intercepts: (-1, 0), (1, 0)
Extrema: local min at (0, -3)
Inflection points: none found in view
Vertical asymptotes: none found in view
Horizontal asymptotes: none found
Domain: all real numbers in view
Range (estimated): approximately [-3, 297]

Tip

Identify the outer and inner functions before differentiating anything — ask 'what's the very last operation applied?' That operation is the outer function; everything inside it is the inner function.

Common Mistakes

Differentiating (3x + 1)⁵ as 5(3x + 1)⁴ and stopping, forgetting to multiply by the inner function's derivative.
The chain rule always has two factors — the outer derivative evaluated at the inner function, and the inner function's own derivative. Dropping the second factor (here, ×3) is the single most common chain rule error.
Multiplying by g(x) itself instead of g'(x).
The chain rule's second factor is the inner function's derivative, g'(x) — not the inner function's value. For g(x) = 3x + 1, that factor is 3, not (3x + 1).

Key Takeaways

The chain rule differentiates a composite function f(g(x)) as f'(g(x)) · g'(x) — outer derivative at the inner function, times inner derivative.
Identifying the outer and inner functions first — by asking what operation is applied last — is the key step before applying the rule.
Rates of change compound by multiplication through a composition, the same intuition behind the chain rule's product of two derivatives.

Summary

The chain rule handles functions wrapped inside functions. The next lesson applies that same tool to equations that aren't solved for y at all.

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