Chapter 2 — Part 2: The Chain Rule

Mathematics Hip-Hop • UACE Pure Mathematics

CHAPTER 02 — DIFFERENTIATION

PART 2: THE CHAIN RULE

Learn how to differentiate a function that contains another function.

LEARN IT. PRACTICE IT. MASTER IT. SHARE IT.

1. Learning objectives

By the end of this lesson, you should be able to:

• recognise a composite function (a function inside another function);
• identify the inner and outer functions;
• state and apply the chain rule correctly;
• differentiate powers of brackets, roots and more complicated expressions;
• combine the chain rule with basic differentiation rules.

2. What is a composite function?

A composite function is formed when one function is placed inside another. For example, in y = (3x + 2)⁵, the expression 3x + 2 is inside the fifth-power operation.

Think of the expression as two layers:

• Inner function: the expression inside the brackets, here u = 3x + 2.
• Outer function: the operation performed on the inner expression, here y = u⁵.

Key idea: Differentiate the outside first, keep the inside unchanged, and then multiply by the derivative of the inside.

3. The chain rule

If y = f(u) and u = g(x), then y is a function of x through u. The chain rule states:

dy/dx = (dy/du) × (du/dx)

Equivalently, if y = f(g(x)):

dy/dx = f′(g(x)) × g′(x)

A particularly useful form is obtained when y = [u(x)]ⁿ:

dy/dx = n[u(x)]ⁿ⁻¹ × du/dx

A reliable four-step method

1. Identify the expression inside the brackets and call it u.
2. Differentiate the outer function with respect to u.
3. Multiply by du/dx.
4. Simplify the final expression carefully.

4. Fully worked examples

Example 1: A linear expression raised to a power

Differentiate y = (3x + 2)⁵.

Step 1 — Identify the inner function:
Let u = 3x + 2. This is the expression inside the brackets.

Step 2 — Differentiate the inner function:
du/dx = d(3x + 2)/dx = 3.

Step 3 — Differentiate the outer function:
Since y = u⁵, differentiate with respect to u:
dy/du = 5u⁴.

Step 4 — Multiply and substitute:
By the chain rule, dy/dx = (dy/du)(du/dx).
dy/dx = 5u⁴ × 3 = 15u⁴.
Now substitute u = 3x + 2.

Answer: dy/dx = 15(3x + 2)⁴

Example 2: A quadratic expression raised to a power

Differentiate y = (2x² − 1)⁴.

Step 1 — Identify the inner function:
Let u = 2x² − 1.

Step 2 — Differentiate the inner function:
du/dx = d(2x² − 1)/dx = 4x.

Step 3 — Differentiate the outer function:
Since y = u⁴, dy/du = 4u³.

Step 4 — Apply the chain rule:
dy/dx = (dy/du)(du/dx) = 4u³ × 4x = 16xu³.
Substitute u = 2x² − 1.

Answer: dy/dx = 16x(2x² − 1)³

Example 3: A square root function

Differentiate y = √(5x − 1).

Step 1 — Rewrite the square root:
y = √(5x − 1) = (5x − 1)^1/2.

Step 2 — Identify and differentiate the inner function:
Let u = 5x − 1.
du/dx = d(5x − 1)/dx = 5.

Step 3 — Differentiate the outer function:
Since y = u^1/2, dy/du = (1/2)u^−1/2.

Step 4 — Apply the chain rule and simplify:
dy/dx = (1/2)u^−1/2 × 5 = (5/2)u^−1/2.
Replace u by 5x − 1 and write the negative power as a square-root denominator.

Answer: dy/dx = 5/[2√(5x − 1)]

For a real-valued function, the original square root requires 5x − 1 ≥ 0. The derivative formula applies where 5x − 1 > 0.

Example 4: A negative power

Differentiate y = (x² + 4)⁻³.

Step 1 — Identify the inner function:
Let u = x² + 4, so that y = u⁻³.

Step 2 — Differentiate the inner function:
du/dx = d(x² + 4)/dx = 2x.

Step 3 — Differentiate the outer function:
dy/du = −3u⁻⁴. The exponent −3 comes down as a multiplier, then decreases by 1.

Step 4 — Apply the chain rule and rewrite:
dy/dx = (−3u⁻⁴)(2x) = −6xu⁻⁴.
Replace u by x² + 4 and write with a positive power in the denominator.

Answer: dy/dx = −6x/(x² + 4)⁴

Example 5: A cubic expression inside a square root

Differentiate y = √(x³ + 2x).

Step 1: Write y = (x³ + 2x)^1/2.

Step 2: Let u = x³ + 2x. Then du/dx = 3x² + 2.

Step 3: dy/du = (1/2)u^−1/2.

Step 4: Multiply:
dy/dx = (1/2)(x³ + 2x)^−1/2(3x² + 2).

Answer: dy/dx = (3x² + 2)/[2√(x³ + 2x)]

The derivative is defined where the expression under the square root is positive.

Example 6: Chain rule combined with a sum

Differentiate y = 3(2x − 1)⁴ + 7.

The constant multiplier 3 remains. The derivative of the constant 7 is zero.

dy/dx = 3 × 4(2x − 1)³ × 2 = 24(2x − 1)³

5. Common mistakes to avoid

• Forgetting the inner derivative: the derivative of (3x + 2)⁵ is not just 5(3x + 2)⁴; it must also be multiplied by 3.
• Changing the bracket too early: keep the inner expression unchanged while differentiating the outer function.
• Dropping negative signs: for u⁻³, the outer derivative is −3u⁻⁴.
• Confusing the power rule and chain rule: when the base is more than x, multiply by the derivative of that base.

Remember: A useful check is to ask, “Have I multiplied by the derivative of everything inside the brackets?”

6. Practice exercises

Differentiate each function with respect to x. Show your steps, identify the inner function where appropriate, and simplify your answers.

1. y = (x + 4)⁶
2. y = (5x − 3)⁴
3. y = (x² + 1)³
4. y = (3x² − 2x)⁵
5. y = √(4x + 7)
6. y = (2x² + 5)⁻²
7. y = 7(3x − 1)³ − 6
8. y = (x³ − 4x + 2)²
9. y = 1/√(x + 3)
10. y = √(2x² − 5x + 6)

7. Answers

Attempt all the questions before opening the answers.

Reveal answers to Exercises 1–10

1. dy/dx = 6(x + 4)⁵
2. dy/dx = 20(5x − 3)³
3. dy/dx = 6x(x² + 1)²
4. dy/dx = 5(3x² − 2x)⁴(6x − 2)
5. dy/dx = 2/√(4x + 7)
6. dy/dx = −8x/(2x² + 5)³
7. dy/dx = 63(3x − 1)²
8. dy/dx = 2(x³ − 4x + 2)(3x² − 4)
9. dy/dx = −1/[2(x + 3)^3/2]
10. dy/dx = (4x − 5)/[2√(2x² − 5x + 6)]

8. Lesson summary

• The chain rule is used when one function is inside another.
• Differentiate the outer function while keeping the inner expression in place.
• Multiply by the derivative of the inner expression.
• Use the chain rule with positive, negative and fractional powers.
• Always simplify carefully and check signs and brackets.

Coming next — Part 3: The Product Rule and Quotient Rule. You will learn how to differentiate products and quotients of functions.

MATHEMATICS HIP-HOP
Learn it. Practice it. Master it. Share it.