Grade by Grade
A. Rate of Change · Derivatives and Differentiation Rules

The Chain Rule

A3.4 A3.5About 70 minutes
Learning goals
  • I can identify the inner and outer functions of a composite function.
  • I can verify the chain rule with monomial functions by simplifying first.
  • I can use the chain rule in function notation and in Leibniz notation.
  • I can differentiate powers and roots of polynomials, and solve problems where rates multiply.

Words to know

composite function
A function made by applying one function to the output of another: f(g(x)), also written (f ∘ g)(x).
inner function
In f(g(x)), the function g that is applied first.
outer function
In f(g(x)), the function f that is applied to the result of the inner function.
chain rule
If y = f(g(x)), then dy/dx = f′(g(x))g′(x); in Leibniz form, dy/dx = (dy/du)(du/dx).

Functions inside functions

How would you differentiate y = (x² + 1)¹⁰? You could expand it into a polynomial with 11 terms… or not. Or y = √(25 − x²)? There is nothing to expand at all.

Both are composite functions: one function applied to the output of another. In MHF4U, unit Combining Functions and Modelling, you wrote compositions as f(g(x)) or (f ∘ g)(x). For y = (x² + 1)¹⁰, the inner function is g(x) = x² + 1 and the outer function is f(u) = u¹⁰.

The chain rule differentiates compositions.

The chain rule

If y = f(g(x)), then

dy/dx = f′(g(x)) · g′(x)

In words: the derivative of the outer function, evaluated at the inner function, times the derivative of the inner function.

In Leibniz notation, with u = g(x) and y = f(u):

dy/dx = (dy/du) · (du/dx)

xinner functionu = g(x) = x² + 1outer functiony = f(u) = u³xudu/dx = 2xdy/du = 3u²dy/dx = (dy/du)(du/dx) = 3u² · 2x = 3(x² + 1)² · 2xy = (x² + 1)³ is a composition: y = f(g(x))
A composite function as two machines in a row. Each machine has its own rate of change, and the overall rate is the product of the two rates: the chain rule.

Why rates multiply

Suppose u changes 3 times as fast as x (du/dx = 3), and y changes 2 times as fast as u (dy/du = 2). Then y changes 2 × 3 = 6 times as fast as x. It is like gears: if the middle gear turns 3 times for each turn of the first, and the last gear turns 2 times for each turn of the middle one, the last gear turns 6 times for each turn of the first.

The Leibniz form dy/du · du/dx looks as if the du's cancel. They are not really fractions, but the notation was designed so that this memory aid gives the right answer.

Worked example

Example 1: Verifying the chain rule with monomials

Verify the chain rule for (a) y = (x²)³ and (b) y = (3x)², by simplifying first and comparing.

  1. (a) Simplify: y = x⁶, so dy/dx = 6x⁵.
  2. Chain rule: outer u³, inner u = x². dy/dx = 3(x²)² · 2x = 3x⁴ · 2x = 6x⁵. ✓
  3. (b) Simplify: y = 9x², so dy/dx = 18x.
  4. Chain rule: outer u², inner u = 3x. dy/dx = 2(3x) · 3 = 18x. ✓

The chain rule agrees with the simplified forms: 6x⁵ and 18x. Notice that without the factor g′(x) (the 2x and the 3) the answers would be wrong.

Worked example

Example 2: A power of a polynomial

Differentiate y = (x² + 1)³. Check by expanding.

  1. Outer: u³ → 3u². Inner: x² + 1 → 2x.
  2. dy/dx = 3(x² + 1)² · 2x = 6x(x² + 1)².
  3. Check: (x² + 1)³ = x⁶ + 3x⁴ + 3x² + 1, so dy/dx = 6x⁵ + 12x³ + 6x.
  4. And 6x(x² + 1)² = 6x(x⁴ + 2x² + 1) = 6x⁵ + 12x³ + 6x. ✓

dy/dx = 6x(x² + 1)². Leave it in this factored form: it is shorter and easier to use (for example, it is clearly 0 only at x = 0).

The power-of-a-function rule

A very common special case: d/dx [g(x)]ⁿ = n[g(x)]ⁿ⁻¹ · g′(x). Bring the power down, keep the inside the same, lower the power by 1, then multiply by the derivative of the inside.

Worked example

Example 3: A radical

Differentiate y = √(25 − x²) and find the slope at (3, 4).

  1. Rewrite: y = (25 − x²)^(1/2).
  2. dy/dx = (1/2)(25 − x²)^(−1/2) · (−2x) = −x/√(25 − x²).
  3. At x = 3: −3/√16 = −3/4.

dy/dx = −x/√(25 − x²); the slope at (3, 4) is −3/4.

xy-555(3, 4)radiusslope 4/3y = √(25 − x²) (upper semicircle)Chain rule: dy/dx = −x/√(25 − x²)At (3, 4): dy/dx = −3/4dashed line: tangent, slope −3/4(4/3)(−3/4) = −1, so the tangentis perpendicular to the radius,a known fact of circle geometry.
The upper half of the circle x² + y² = 25, drawn with equal scales. The chain rule gives the tangent slope −3/4 at (3, 4), and the tangent is perpendicular to the radius, which has slope 4/3.
Worked example

Example 4: A power in the denominator

Differentiate f(x) = 1/(2x − 5)⁴.

  1. Rewrite: f(x) = (2x − 5)⁻⁴.
  2. f′(x) = −4(2x − 5)⁻⁵ · 2 = −8(2x − 5)⁻⁵.
  3. With a positive exponent: f′(x) = −8/(2x − 5)⁵.

f′(x) = −8/(2x − 5)⁵ (for x ≠ 5/2).

Worked example

Example 5: Rates that multiply

A pebble dropped in a calm lake makes a circular ripple whose radius grows at 30 cm/s, so r = 30t. How fast is the area inside the ripple growing at t = 2 s?

  1. A = πr², so dA/dr = 2πr. And dr/dt = 30.
  2. Chain rule: dA/dt = (dA/dr)(dr/dt) = 2πr · 30.
  3. At t = 2, r = 60 cm: dA/dt = 2π(60)(30) = 3600π ≈ 11 310 cm²/s.

The area grows at 3600π ≈ 11 310 cm²/s (about 1.13 m² every second). You will solve many more problems like this (related rates) in Applications of Derivatives.

Just for fun

A student once complained that the chain rule was "just the power rule with extra steps." Their teacher replied that, with a long enough chain of inner functions, it is the power rule with extra steps, extra steps and extra steps.

Hands-on activity

Gear-ratio chain

You need: a bicycle with gears (with a grown-up's help; keep fingers away from the chain), or three round lids of different sizes and a marker.

Bicycle version: Turn the pedals slowly by hand through one full turn with the back wheel lifted, and count how many times the back wheel turns. That ratio is d(wheel turns)/d(pedal turns). Now find d(distance)/d(wheel turns) by measuring the wheel's circumference with a string. Multiply the two rates to get distance per pedal turn.

Lid version: Mark a dot on the edge of each lid. Roll lid A along a line until it has turned once, and measure the distance (its circumference). Do the same for B and C. If A drives B by rolling against it, B turns (circumference A ÷ circumference B) times per turn of A. Predict C's turns per turn of A by multiplying the two ratios.

What to look for: on many bikes in a middle gear the back wheel turns about 2 to 3 times per pedal turn, and a 66 cm wheel has a circumference near 2.07 m, so one pedal turn moves the bike roughly 4 to 6 m. In both versions the overall rate is the product of the rates in the chain.

Your turn

1

Identify the inner and outer functions of y = (3x − 1)⁵.

2

Differentiate y = (3x − 1)⁵.

3

Differentiate f(x) = (x³ + 2x)².

4

Differentiate y = √(4x + 1).

5

Differentiate g(x) = 1/(x² + 1).

6

Verify the chain rule for y = (x³)² by simplifying first.

7

If y = u⁴ and u = 2x + 7, find dy/dx when x = −3.

More practice: warm-up

1

d/dx (x + 1)² = ?

2

d/dx (5x)³ = ?

3

What is the inner function of √(x² + 9)?

4

d/dx (2x + 3)⁴ = ?

5

Write the chain rule in Leibniz notation.

More practice: on your own

1

Differentiate y = (1 − x²)⁶.

2

Differentiate f(x) = ∛(x² − 8) and evaluate f′(4).

3

Find the equation of the tangent to y = (x² − 3)⁴ at x = 2.

4

Find the points where y = (x² − 4)² has a horizontal tangent.

5

A spherical balloon's radius grows at 2 cm/s. How fast is its volume growing when r = 10 cm? (V = (4/3)πr³)

Challenge

1

Differentiate y = ((x² + 1)³ + 2)⁴, a chain of three functions.

2

Given f(2) = 5, f′(2) = 3, g(5) = 1 and g′(5) = −4, find the derivative of g(f(x)) at x = 2.

3

Show that the tangent to the circle x² + y² = r² at any point (a, b) on the upper half (b > 0) is perpendicular to the radius.

Check your understanding

Quiz

1.

What is the derivative of y = (4x + 1)³?

2.

Which statement is the chain rule?

3.

For y = √(x² + 9), what is dy/dx at x = 4?

4.

What is d/dx [1/(x − 3)²]?

5.

A student differentiates (x² + 5)³ as 3(x² + 5)². What did they forget?

6.

If f(1) = 2, f′(1) = 4 and g′(2) = −3, what is the derivative of g(f(x)) at x = 1?

For parents and teachers

Big idea (A3.4, A3.5): to differentiate a function inside another function, multiply the derivative of the outer function (with the inside left alone) by the derivative of the inside. Rates along a chain multiply.

How to help: Ask your teen to name the "inside" and the "outside" before differentiating. For y = (x² + 1)³: inside x² + 1, outside "cube it". Have them check simple cases by expanding, as in Example 1.

Common difficulties:

  • Forgetting to multiply by the inner derivative (by far the most common error).
  • Differentiating the inside within the brackets: writing 3(2x)² instead of 3(x² + 1)² · 2x.
  • Losing negative signs with negative exponents: (2x − 5)⁻⁴ becomes −4(2x − 5)⁻⁵ · 2.
  • Confusing composition f(g(x)) with multiplication f(x)g(x).