The Chain Rule: Study Guide (Calculus AB 3.1)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so derivatives are written in a compact form:

• f′(x) is the derivative of f with respect to x.
• dy/dx is the derivative of y with respect to x. d/dx [ … ] means “the derivative of the bracket with respect to x”.
• f(g(x)) is a composite function: first apply g to x, then apply f to the result. It is also written (f ∘ g)(x).
• sin³x means (sin x)³. e^(1 − x²) means e raised to the power 1 − x².

Why a new rule is needed

Take y = (3x + 1)². You could expand first: y = 9x² + 6x + 1, so dy/dx = 18x + 6 = 6(3x + 1).

Now try a shortcut that only uses the power rule: “bring down the 2 and lower the power”, giving 2(3x + 1). That is wrong. It is too small by a factor of 3.

Where does the 3 come from? The bracket 3x + 1 changes 3 times as fast as x. The square then changes 2(3x + 1) times as fast as the bracket. So y changes 2(3x + 1) × 3 times as fast as x. The rates multiply.

Expanding worked here, but it fails for (3x + 1)⁵⁰ (too long) and for √(3x + 1) or sin(3x + 1) (there is nothing to expand). The chain rule handles all of these in one step.

Outer and inner functions

A composite function is a function applied to the output of another function. To use the chain rule you must first split it into two parts:

• the inner function u = g(x): what you would calculate first if someone gave you a value of x;
• the outer function f(u): what you do to that result last.

Function h(x)	Outer f(u)	Inner u = g(x)	h′(x)
√(1 + 5x)	√u	1 + 5x	5/(2√(1 + 5x))
e^(1 − x²)	eᵘ	1 − x²	−2x e^(1 − x²)
cos(πx)	cos u	πx	−π sin(πx)
ln(sin x), for sin x > 0	ln u	sin x	cos x / sin x = cot x
tan³x	u³	tan x	3 tan²x sec²x
1/(x² + 1)	u⁻¹	x² + 1	−2x/(x² + 1)²

Watch the last two rows closely. tan³x is a power of tan x, so the outer function is “cube” and the inner is tan x. 1/(x² + 1) needs no quotient rule: write it as (x² + 1)⁻¹ and use the chain rule.

The chain rule

Chain rule. If g is differentiable at x and f is differentiable at g(x), then h(x) = f(g(x)) is differentiable at x and h′(x) = f′(g(x)) · g′(x)

Read it in three moves:

1. Differentiate the outer function, f′.
2. Keep the inside exactly as it was: evaluate f′ at g(x), not at x.
3. Multiply by the derivative of the inside, g′(x).

The phrase to say every time is: “derivative of the outside, inside left alone, times the derivative of the inside.”

Leibniz form and units

Write y = f(u) and u = g(x). Then

dy/dx = (dy/du) · (du/dx)

This looks like cancelling du, and that is a good memory aid (it is not a proof). It also explains units. Suppose, in a fictional greenhouse model, relative humidity H (in percentage points) depends on temperature T (in °C), and T depends on time t (in hours). At one moment:

• dH/dT = −1.5 percentage points per °C,
• dT/dt = 2 °C per hour.

Then dH/dt = (dH/dT) · (dT/dt) = (−1.5)(2) = −3 percentage points per hour. The units show why: (points/°C) × (°C/hour) = points/hour.

One detail matters: dH/dT must be the rate at the temperature the greenhouse has at that moment. That is the Leibniz version of “evaluate f′ at g(x)”.

More than two layers

Some functions have three or more layers. Work from the outside in, and write one factor for each layer.

Example: y = [ln(2x + 1)]³. The layers are “cube”, then “ln”, then “2x + 1”.

• Cube: 3[ln(2x + 1)]²
• times ln: 1/(2x + 1)
• times 2x + 1: 2

So dy/dx = 3[ln(2x + 1)]² · 1/(2x + 1) · 2 = 6[ln(2x + 1)]²/(2x + 1).

Each factor’s inside is still left alone. Stop when the innermost function has been differentiated; for 2x + 1 that factor is just 2.

Worked example 1: a square root and a tangent line

Question. Let f(x) = √(x² + 9). Find f′(x), then the equation of the tangent line to the graph of f at x = 4.

1. Identify the layers. Outer: √u = u^(1/2). Inner: u = x² + 9.
2. Differentiate the outer function, inside unchanged. d/du [u^(1/2)] = (1/2)u^(−1/2), so this factor is 1/(2√(x² + 9)).
3. Multiply by the derivative of the inside. d/dx [x² + 9] = 2x. f′(x) = 2x/(2√(x² + 9)) = x/√(x² + 9)
4. Evaluate at x = 4. f(4) = √(16 + 9) = √25 = 5, and f′(4) = 4/5.
5. Write the tangent line through (4, 5) with slope 4/5: y = 5 + (4/5)(x − 4), which is y = (4/5)x + 9/5.

Check. (f(4.001) − f(4))/0.001 ≈ 0.80004, which is close to 4/5 = 0.8.

Common slip. Writing f′(x) = 1/(2√(x² + 9)) forgets the factor 2x. That answer would give slope 1/10 at x = 4, which is far too small.

Worked example 2: the chain rule from a table

Questions often give only values, not formulas. The table shows values of two differentiable functions f and g.

x	f(x)	f′(x)	g(x)	g′(x)
1	3	−2	2	5
2	4	6	3	−1
3	1	7	1	4

(a) h(x) = f(g(x)). Find h′(1).

1. Chain rule: h′(1) = f′(g(1)) · g′(1).
2. Inside first: g(1) = 2. So the outer derivative is f′(2) = 6.
3. Derivative of the inside: g′(1) = 5.
4. h′(1) = 6 × 5 = 30.

Two tempting errors: f′(1) · g′(1) = (−2)(5) = −10 evaluates f′ at the wrong input. And f′(g′(1)) = f′(5) is not even in the table, which is a strong hint that it is the wrong expression.

(b) k(x) = g(f(x)). Find k′(3).

The order is reversed, so the roles swap: k′(3) = g′(f(3)) · f′(3) = g′(1) · 7 = 5 × 7 = 35.

(c) p(x) = [g(x)]². Find p′(2).

Outer: u², inner: g(x). p′(2) = 2g(2) · g′(2) = 2(3)(−1) = −6. Writing 2g′(2) = −2 confuses “the derivative of the square” with “the square of the derivative”.

Interpretation. In (a), h′(1) = 30 means that near x = 1, h(x) increases about 30 units for each unit increase in x.

Worked example 3: chain rule inside a product

Question. Let f(x) = x√(4 − x²) for −2 < x < 2. Find f′(x) and the x-values where the tangent line is horizontal.

1. Classify. f is a product: x times √(4 − x²). The second factor is a composite. So use the product rule, and the chain rule inside it.
2. Product rule: f′(x) = (1) · √(4 − x²) + x · d/dx [√(4 − x²)].
3. Chain rule for the second factor. Outer √u, inner 4 − x²: d/dx [√(4 − x²)] = 1/(2√(4 − x²)) · (−2x) = −x/√(4 − x²).
4. Combine: f′(x) = √(4 − x²) − x²/√(4 − x²).
5. Write over one denominator: f′(x) = ((4 − x²) − x²)/√(4 − x²) = (4 − 2x²)/√(4 − x²).
6. Horizontal tangent where f′(x) = 0. The denominator is positive on −2 < x < 2, so set the top to 0: 4 − 2x² = 0, so x² = 2 and x = √2 or x = −√2.

Check. f(√2) = √2 · √2 = 2 and f(−√2) = −2, so the horizontal tangents are at (√2, 2) and (−√2, −2). As a spot check, f′(1) = 2/√3 ≈ 1.15 is positive, which fits a graph that is still rising before x = √2 ≈ 1.41.

Order of operations. Decide the last operation first. Here the last step in computing f(x) is “multiply”, so the product rule is the outer structure and the chain rule sits inside it.

Common misconceptions

• Forgetting the derivative of the inside. d/dx [sin(5x)] is 5 cos(5x), not cos(5x). This is the most frequent error in the whole topic.
• Changing the inside when differentiating the outside. d/dx [cos(x²)] is −sin(x²) · 2x, not −sin(2x) or −sin(2x) · 2x. The inside stays x² in the outer factor.
• Evaluating f′ at x instead of at g(x). In tables, h′(a) = f′(g(a)) · g′(a). Find g(a) first.
• Missing a hidden composite. sin²x, e^(4x), ln(3x) and 1/(x² + 1) are all composites. Ask “is the input exactly x?”
• Stopping too early, or too late. With e^(1 − x²), the derivative of the outer eᵘ is eᵘ itself, kept as e^(1 − x²). Multiply by −2x once and stop. Do not differentiate e^(1 − x²) a second time.
• Mixing up sin²x and sin(x²). d/dx [sin²x] = 2 sin x cos x; d/dx [sin(x²)] = 2x cos(x²). Different inner functions, different answers.
• Using the product rule on a composite. (3x + 1)⁵ is not a product of two different functions; it is a power of one inner function.
• Cancelling du as if it proves the rule. dy/du · du/dx is a helpful pattern, but dy/dx is not a fraction. Use the pattern to remember the rule and track units.

Where this leads

The chain rule is behind almost everything in the rest of Unit 3. Topic 3.2, implicit differentiation, uses it every time you differentiate a term containing y. Topics 3.3 and 3.4 use it to find derivatives of inverse functions, and Unit 4 uses it for related rates. In Unit 6 you will reverse it when you integrate by substitution. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.