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Year 11 Maths

The Chain Rule

Master the technique for differentiating composite functions using the chain rule: dy/dx = dy/du × du/dx.

What Is the Chain Rule?

The chain rule is used to differentiate composite functions -- functions where one function is applied inside another. If y = f(g(x)), the chain rule tells us how to find dy/dx.

The formula is: dy/dx = dy/du × du/dx, where we let u = g(x) be the "inner function" and y = f(u) be the "outer function".

How It Works

Example: Differentiate y = (3x + 1)5

Step 1: Identify inner function

Let u = 3x + 1 (inside the brackets)

Step 2: Rewrite as outer function

Then y = u5 (the outer power)

Step 3: Differentiate each

dy/du = 5u4 and du/dx = 3

Step 4: Multiply together

dy/dx = 5u4 × 3 = 15(3x + 1)4

When to Use the Chain Rule

Use the chain rule whenever you need to differentiate a function of a function. Common situations include:

Powers of expressions

y = (2x2 - 5)7

Square roots of expressions

y = √(4x + 3) = (4x + 3)1/2

Trig of expressions

y = sin(x2)

Exponentials of expressions

y = e3x+1

The key indicator is that there is a "function inside a function". If you can identify an inner part and an outer part, the chain rule applies.

Function Notation Form

In function notation, if y = f(g(x)), then:

dy/dx = f'(g(x)) × g'(x)

This means: differentiate the outer function (keeping the inner function unchanged), then multiply by the derivative of the inner function. This is sometimes described as "differentiate the outside, then multiply by the derivative of the inside."

Key Vocabulary

Composite Function

A function formed by applying one function to the result of another, written as f(g(x)) or (f ∘ g)(x).

Inner Function

The function inside another function, usually called u = g(x). It is the part you substitute first.

Outer Function

The function applied to the inner function, written as y = f(u). You differentiate this with respect to u.

Chain Rule

The rule that states dy/dx = dy/du × du/dx for differentiating composite functions.

Worked Examples

1

Differentiate y = (2x - 7)4

Step 1: Let u = 2x - 7, so y = u4.

Step 2: dy/du = 4u3 and du/dx = 2.

Step 3: dy/dx = dy/du × du/dx = 4u3 × 2 = 8u3.

Answer: dy/dx = 8(2x - 7)3.

2

Differentiate y = √(5x2 + 1)

Step 1: Rewrite as y = (5x2 + 1)1/2. Let u = 5x2 + 1, so y = u1/2.

Step 2: dy/du = (1/2)u-1/2 and du/dx = 10x.

Step 3: dy/dx = (1/2)u-1/2 × 10x = 5x × (5x2 + 1)-1/2.

Answer: dy/dx = 5x / √(5x2 + 1).

3

Differentiate y = e3x+2

Step 1: Let u = 3x + 2, so y = eu.

Step 2: dy/du = eu and du/dx = 3.

Step 3: dy/dx = eu × 3 = 3eu.

Answer: dy/dx = 3e3x+2.

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

If y = (4x + 3)6, what is dy/dx?

Question 2

What is the derivative of y = (x2 + 3x)3?

Question 3

Differentiate y = e5x.

Question 4

If y = (1 - x2)1/2, what is dy/dx?

Question 5

What is the derivative of y = sin(3x)?

Key Concepts Summary

Introduction to Calculus The Product Rule