Practice Maths

The Chain Rule

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Key Terms

A composite function is a function of a function: y = f(g(x)). The inner function is g(x) and the outer function is f.
The chain rule states: if y = f(g(x)), then dy/dx = f′(g(x)) × g′(x).
Equivalently, if y = f(u) and u = g(x), then dy/dx = (dy/du) × (du/dx).
In words: differentiate the outer function (leaving the inner unchanged), then multiply by the derivative of the inner function.
Recognition: look for a function raised to a power, or a polynomial inside another function.
📚 QCAA Formula
Chain rule: dy/dx = f′(g(x)) · g′(x)
Or using substitution u = g(x): dy/dx = (dy/du)(du/dx)
Function y = Inner u = dy/dx =
(2x + 1)52x + 15(2x + 1)4 × 2 = 10(2x + 1)4
(x2 + 3)4x2 + 34(x2 + 3)3 × 2x = 8x(x2 + 3)3
√(3x − 1)3x − 1½(3x − 1)−1/2 × 3 = 3/(2√(3x − 1))
(x3 + 2x)6x3 + 2x6(x3 + 2x)5 × (3x2 + 2)
x inner: g(x) u = g(x) outer: f(u) y = f(g(x)) dy/dx = f'(g(x)) × g'(x)
Hot Tip Write y = (inner)n, identify the inner function u, then apply: bring the power down, reduce the power by 1, multiply by du/dx. Don't forget that last multiplication step — it's the most common error.

Worked Example 1 — Power of a linear function

Question: Differentiate y = (3x + 2)7.

Identify: Outer function f(u) = u7, inner function u = 3x + 2.

Apply chain rule: dy/dx = 7u6 × du/dx = 7(3x + 2)6 × 3

Answer: dy/dx = 21(3x + 2)6

Worked Example 2 — Power of a quadratic

Question: Differentiate y = (x2 − 5)4.

Identify: Outer f(u) = u4, inner u = x2 − 5, so du/dx = 2x.

Apply chain rule: dy/dx = 4(x2 − 5)3 × 2x

Answer: dy/dx = 8x(x2 − 5)3

Worked Example 3 — Square root (fractional power)

Question: Differentiate y = √(4x2 + 1).

Rewrite: y = (4x2 + 1)1/2. Inner u = 4x2 + 1, du/dx = 8x.

Apply chain rule: dy/dx = ½(4x2 + 1)−1/2 × 8x = 4x(4x2 + 1)−1/2

Answer: dy/dx = 4x / √(4x2 + 1)

Introduction

The rules you have learned so far — the power rule, basic polynomial differentiation — apply to simple functions. But many real-world expressions are composed of one function inside another: (2x+1)5, √(x2+3), 1/(x−4)3. The chain rule is the key technique for differentiating these composite functions.

Understanding Composite Functions

A composite function y = f(g(x)) means: first apply g to x, then apply f to the result. For example, y = (2x+1)5 means: first compute u = 2x+1 (inner), then raise to the 5th power (outer).

The chain rule tells us how the rates of change multiply: the rate at which y changes with respect to x equals the rate y changes with u, times the rate u changes with x.

Applying the Chain Rule: Step-by-Step

  1. Identify the inner and outer functions. Write y = f(u) where u = g(x).
  2. Differentiate the outer function with respect to u: find dy/du = f′(u).
  3. Differentiate the inner function with respect to x: find du/dx = g′(x).
  4. Multiply: dy/dx = (dy/du) × (du/dx) = f′(u) × g′(x).
  5. Substitute back u = g(x) to express the answer in terms of x only.
Worked Example 1 — Chain rule with a cubic inner function

Find dy/dx for y = (x3 + 2x + 1)5.

Step 1: Let u = x3 + 2x + 1, so y = u5.

Step 2: dy/du = 5u4.

Step 3: du/dx = 3x2 + 2.

Step 4: dy/dx = 5u4 × (3x2 + 2) = 5(x3 + 2x + 1)4(3x2 + 2).

Negative and Fractional Powers

The chain rule applies to any power, including negative and fractional. Always rewrite square roots as power 1/2 and fractions as negative powers before differentiating.

For example: 1/(x2+1)3 = (x2+1)−3. Apply chain rule: dy/dx = −3(x2+1)−4 × 2x = −6x(x2+1)−4.

Worked Example 2 — Negative power

Differentiate y = 1/(2x − 3)4.

Rewrite: y = (2x − 3)−4.

Chain rule: dy/dx = −4(2x − 3)−5 × 2 = −8(2x − 3)−5.

Answer: dy/dx = −8/(2x − 3)5.

Applications: Rates of Change

The chain rule also appears in related rates problems. If a quantity depends on time t through an intermediate variable, the chain rule links the rates: for example, if V depends on r and r depends on t, then dV/dt = (dV/dr)(dr/dt).

💡 Key Reminder: When you differentiate a composite function, you must always multiply by the derivative of the inner function. Forgetting this step is the single most common chain rule error. A quick check: if the inner function is just x (e.g., y = x5), the derivative of the inner is 1, so the chain rule just gives the power rule — consistent!

Summary

Chain rule: dy/dx = f′(g(x)) × g′(x). Steps: identify inner u = g(x) and outer y = f(u), differentiate each, multiply. Applies to all powers — positive, negative, fractional. Always substitute u back to express in terms of x.

Mastery Practice

See Answers ➔
  1. Fluency

    Differentiate each function using the chain rule.

    1. (a) y = (x + 4)6
    2. (b) y = (3x − 1)4
    3. (c) y = (2x + 5)3
    4. (d) y = (5 − x)7
  2. Fluency

    Differentiate each function. Write your answer in simplified form.

    1. (a) y = (x2 + 1)5
    2. (b) y = (x2 − 4x)3
    3. (c) y = (2x2 + x)4
    4. (d) y = (x3 + 1)6
  3. Fluency

    Differentiate each function involving square roots. Leave answers with positive indices.

    1. (a) y = √(2x + 3)
    2. (b) y = √(x2 + 5)
    3. (c) y = √(4 − x)
    4. (d) y = √(3x2 − x)
  4. Fluency

    Rewrite with negative indices then differentiate. Express the final answer with positive indices.

    1. (a) y = 1/(x + 2)3
    2. (b) y = 1/(2x − 1)5
    3. (c) y = 4/(x2 + 3)2
  5. Understanding

    Find the gradient of each curve at the given point.

    Method: Differentiate using the chain rule, then substitute the x-value.
    1. (a) y = (2x + 1)4 at x = 0
    2. (b) y = (x2 + 3)3 at x = 1
    3. (c) y = √(x + 5) at x = 4
    4. (d) y = (3 − x)5 at x = 2
  6. Understanding

    Find the equation of the tangent to each curve at the given point.

    1. (a) y = (x + 1)4 at x = 1
    2. (b) y = (2x − 3)3 at x = 2
    3. (c) y = √(2x + 1) at x = 4
  7. Understanding

    Differentiate these more complex composite functions.

    1. (a) y = (x2 + 2x + 1)4
    2. (b) y = (x3 − 3x)5
    3. (c) y = (1 − 2x2)3
    4. (d) y = ((x + 1)2)3   [Hint: simplify first, then differentiate]
  8. Understanding

    Mixed differentiation — chain rule and power rule together.

    Find f′(x) for each function.

    1. (a) f(x) = 3x4 + 2(x + 1)5
    2. (b) f(x) = (x2 + 1)3 − x3
    3. (c) f(x) = 5(2x − 1)4 + 3x
  9. Problem Solving

    Related rates using the chain rule.

    Challenge. A spherical balloon is being inflated. Its radius at time t seconds is r = (3t + 1) cm.
    1. (a) Write an expression for the volume V of the sphere in terms of t. (Use V = (4/3)πr3.)
    2. (b) Use the chain rule (dV/dt = (dV/dr)(dr/dt)) to find dV/dt.
    3. (c) Find the rate of change of volume when t = 1 second. Leave your answer in terms of π.
  10. Problem Solving

    Chain rule applied to motion.

    Challenge. A particle moves along a line. Its displacement (in metres) at time t seconds is given by s = (t2 + 2)3.
    1. (a) Find the velocity v = ds/dt at time t.
    2. (b) Find the velocity when t = 1 second.
    3. (c) Find the acceleration a = dv/dt at time t = 1. (Hint: differentiate your velocity expression using the chain rule again.)
  11. Problem Solving

    Stationary points using the chain rule.

    For the curve y = (2x − 4)4 − 3:

    1. (a) Find dy/dx using the chain rule.
    2. (b) Find all stationary points (where dy/dx = 0).
    3. (c) Determine the nature of each stationary point.