AP Physics C: Mechanics · Unit 0: Calculus Primer · Lesson 0.3

Deep Dive: The Chain Rule

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
Unit 0 · 0.3Concept

What Is a Composite Function?

A composite function is one function tucked inside another — an "outside" function acting on the result of an "inside" function. Two examples you'll see constantly this year:

  • h(t) = (3t + 1)² — the outside function is "square it," the inside function is 3t + 1
  • x(t) = sin(ωt + φ) — the outside function is "take the sine," the inside function is ωt + φ

In function notation, a composite is written h(t) = f(g(t)), where g(t) is the inside function and f(u) is the outside function acting on whatever g(t) produces. The power rule alone can differentiate f or g separately — but not the two of them wired together. That's the job of the chain rule.

Unit 0 · 0.3Math

The Chain Rule

To differentiate a composite function, differentiate the outside function first — leaving whatever is inside completely untouched — then multiply by the derivative of the inside function:

d/dt f(g(t)) = f'(g(t)) · g'(t)

In plain language: derivative of the outside, times derivative of the inside. One more fact you'll need before applying this to physics: the derivatives of sine and cosine.

d/dx(sin x) = cos x
d/dx(cos x) = −sin x
💡These two facts are given here as known results, not derived from the limit definition — that derivation involves a couple of trig limits that are outside the scope of this primer. What matters for this course is being able to usethem correctly inside the chain rule.
Unit 0 · 0.3Math

Worked Example — Checking the Chain Rule

Before trusting the chain rule on something you can't easily double-check (like a trig function), it's worth verifying it against a case you can check — by expanding the composite function directly and differentiating the ordinary way.

ExampleWorked Example — Differentiating h(t) = (3t + 1)² Two Ways

Find h'(t) for h(t) = (3t + 1)², using the chain rule, then verify the result by expanding first.

🔑Both methods agree — which is exactly what should happen. Expanding first works fine for a simple square like this one, but it becomes unmanageable fast (try expanding (3t + 1)¹⁰ by hand). The chain rule gets the same correct answer without ever needing to expand anything.
Unit 0 · 0.3Example

Differentiating a Trig Composite

Expanding isn't an option for a trig composite — there's no algebra trick that unpacks sin(ωt + φ) into separate pieces. The chain rule is the only way in.

ExampleGuided Example — Differentiating x(t) = A sin(ωt + φ)

Find dx/dt for x(t) = A sin(ωt + φ), where A, ω, and φ are constants.

Step 1Identify inside and outside
Inside: g(t) = ωt + φ, so g'(t) = ω. Outside: f(u) = A sin(u), so f'(u) = A cos(u).
Unit 0 · 0.3Math

Capstone — Verifying the SHM Differential Equation

Here's the payoff, and it's a big one for later this year. Unit 7 will hand you the equation x(t) = A cos(ωt + φ) and ask you to recognize it as the solution to a specific differential equation. Right now, with nothing but the chain rule, you can prove it's true.

ExampleWorked Example — Differentiating x(t) = A cos(ωt + φ) Twice

Given x(t) = A cos(ωt + φ), find v(t) = dx/dt and a(t) = d²x/dt², and compare a(t) to x(t).

Explore this relationship directly below — no matter how you set A, ω, and φ, the acceleration graph always comes out as an upside-down, rescaled copy of the position graph.

Adjust amplitude A, angular frequency ω, and phase φ for x(t) = A cos(ωt + φ). Each derivative below was found using the chain rule — watch how a(t) always looks like an upside-down, rescaled copy of x(t).

A (amplitude) = 2.00
ω (angular frequency) = 1.00
φ (phase, radians) = 0.00
x(t) = 2.0·cos(1.00t + 0.00)v(t) = dx/dt = −2.00·sin(1.00t + 0.00)a(t) = d²x/dt² = −2.00·cos(1.00t + 0.00) = −ω²·x(t)
x(t) — position
2.4-2.4t=10s
meters
v(t) — velocity
2.4-2.4t=10s
m/s
a(t) — acceleration
2.4-2.4t=10s
m/s²

a(t)'s graph is x(t)'s graph, flipped upside down and rescaled by ω². That's exactly what d²x/dt² = −ω²x means — acceleration is always proportional to position, and always points back the opposite way.

🔑Lesson 7.3 will ask you to recognize that x(t) = A cos(ωt + φ) solves d²x/dt² = −ω²x — not derive it from scratch. You now know exactly why that recognition is justified.
← Back to Lesson 0.3Ready for 0.4? The Power Rule for Integrals shows you how to run all of this in reverse.