CHAPTER 02 — DIFFERENTIATION | PART 3: PRODUCT RULE AND QUOTIENT RULE

Mathematics Hip-Hop • UACE Pure Mathematics

CHAPTER 02 — DIFFERENTIATION

PART 3: PRODUCT RULE AND QUOTIENT RULE

Learn how to differentiate functions formed by multiplication and division.

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1. Learning objectives

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

• recognise when a function is a product or a quotient of two functions;
• identify the functions represented by u and v;
• state and apply the product rule accurately;
• state and apply the quotient rule accurately;
• combine these rules with the chain rule where necessary;
• simplify derivatives and state any restrictions on the variable.

2. The product rule

Use the product rule when y is the product of two functions of x. Suppose

y = u × v

where both u and v depend on x. The derivative is:

dy/dx = u(dv/dx) + v(du/dx)

In prime notation:

(uv)′ = uv′ + vu′

Remember: Keep the first function and differentiate the second, then keep the second function and differentiate the first. Add the two results.

Step-by-step method

Step 1: Identify the two factors and name them u and v.

Step 2: Differentiate each factor separately to find u′ and v′.

Step 3: Substitute into uv′ + vu′.

Step 4: Expand and simplify the result.

3. Worked examples: product rule

Example 1: Differentiate y = (x² + 1)(3x − 2).

Step 1: Identify the factors.

Let u = x² + 1 and v = 3x − 2.

Step 2: Differentiate each factor.

u′ = 2x and v′ = 3.

Step 3: Apply the product rule.

dy/dx = uv′ + vu′

= (x² + 1)(3) + (3x − 2)(2x).

Step 4: Expand and simplify.

= 3x² + 3 + 6x² − 4x

Final answer: dy/dx = 9x² − 4x + 3.

Example 2: Differentiate y = x³ sin x.

Step 1: Let u = x³ and v = sin x.

Step 2: u′ = 3x² and v′ = cos x.

Step 3: dy/dx = uv′ + vu′

= x³ cos x + (sin x)(3x²).

Step 4: Arrange the terms neatly.

Final answer: dy/dx = x³ cos x + 3x² sin x.

Example 3: Differentiate y = (2x + 1)eˣ.

Step 1: Let u = 2x + 1 and v = eˣ.

Step 2: u′ = 2 and v′ = eˣ.

Step 3: dy/dx = uv′ + vu′

= (2x + 1)eˣ + 2eˣ.

Step 4: Factor out eˣ.

= eˣ[(2x + 1) + 2].

Final answer: dy/dx = eˣ(2x + 3).

Example 4: Differentiate y = x² ln x, where x > 0.

Step 1: Let u = x² and v = ln x.

Step 2: u′ = 2x and v′ = 1/x.

Step 3: dy/dx = uv′ + vu′

= x²(1/x) + (ln x)(2x).

Step 4: Simplify.

= x + 2x ln x.

Final answer: dy/dx = x(1 + 2 ln x).

4. The quotient rule

Use the quotient rule when one function is divided by another. Let

y = u/v, where v ≠ 0

The quotient rule is:

dy/dx = [v(du/dx) − u(dv/dx)] / v²

In prime notation:

(u/v)′ = (vu′ − uv′)/v²

Important order: Bottom × derivative of top, MINUS top × derivative of bottom, all divided by bottom squared.

Step-by-step method

Step 1: Identify the numerator u and denominator v.

Step 2: Differentiate them separately to obtain u′ and v′.

Step 3: Substitute into (vu′ − uv′)/v². Keep the order of the subtraction correct.

Step 4: Expand, simplify and state where the original denominator is not zero.

5. Worked examples: quotient rule

Example 5: Differentiate y = (x + 1)/(x − 2).

Step 1: Let u = x + 1 and v = x − 2.

Step 2: u′ = 1 and v′ = 1.

Step 3: dy/dx = (vu′ − uv′)/v²

= [(x − 2)(1) − (x + 1)(1)]/(x − 2)².

Step 4: Simplify the numerator.

= (x − 2 − x − 1)/(x − 2)²

Final answer: dy/dx = −3/(x − 2)², x ≠ 2.

Example 6: Differentiate y = (x² + 1)/(2x + 3).

Step 1: Let u = x² + 1 and v = 2x + 3.

Step 2: u′ = 2x and v′ = 2.

Step 3: dy/dx = (vu′ − uv′)/v²

= [(2x + 3)(2x) − (x² + 1)(2)]/(2x + 3)².

Step 4: Expand the numerator.

= (4x² + 6x − 2x² − 2)/(2x + 3)².

Final answer: dy/dx = (2x² + 6x − 2)/(2x + 3)², x ≠ −3/2.

Example 7: Differentiate y = eˣ/x.

Step 1: Let u = eˣ and v = x.

Step 2: u′ = eˣ and v′ = 1.

Step 3: dy/dx = (vu′ − uv′)/v²

= [xeˣ − eˣ(1)]/x².

Step 4: Factorise the numerator.

Final answer: dy/dx = eˣ(x − 1)/x², x ≠ 0.

Example 8: Differentiate y = (3x − 1)/(x² + 1).

Step 1: Let u = 3x − 1 and v = x² + 1.

Step 2: u′ = 3 and v′ = 2x.

Step 3: dy/dx = (vu′ − uv′)/v²

= [3(x² + 1) − (3x − 1)(2x)]/(x² + 1)².

Step 4: Expand and simplify.

= (3x² + 3 − 6x² + 2x)/(x² + 1)².

Final answer: dy/dx = (−3x² + 2x + 3)/(x² + 1)².

6. Combining the rules

Some functions require more than one differentiation rule. Decide which rule applies to the overall structure, then use the chain rule inside a factor when necessary.

Example 9: Differentiate y = x²(3x + 1)⁴.

Step 1: This is a product. Let u = x² and v = (3x + 1)⁴.

Step 2: u′ = 2x. For v, use the chain rule: v′ = 4(3x + 1)³ × 3 = 12(3x + 1)³.

Step 3: Apply the product rule.

dy/dx = x²[12(3x + 1)³] + (3x + 1)⁴(2x).

Step 4: Factorise the common terms.

= 2x(3x + 1)³[6x + (3x + 1)].

Final answer: dy/dx = 2x(3x + 1)³(9x + 1).

Example 10: Differentiate y = (x² + 1)³/x.

Step 1: This is a quotient. Let u = (x² + 1)³ and v = x.

Step 2: Use the chain rule for u: u′ = 3(x² + 1)² × 2x = 6x(x² + 1)². Also v′ = 1.

Step 3: Apply the quotient rule.

dy/dx = [x·6x(x² + 1)² − (x² + 1)³]/x².

Step 4: Factorise.

= (x² + 1)²[6x² − (x² + 1)]/x².

Final answer: dy/dx = (x² + 1)²(5x² − 1)/x², x ≠ 0.

7. Common mistakes to avoid

• Multiplying the derivatives: The product rule is not u′v′. It is uv′ + vu′.
• Reversing the quotient numerator: Use vu′ − uv′, not uv′ − vu′.
• Forgetting to square the denominator: The quotient-rule denominator is v².
• Forgetting the chain rule: For example, the derivative of (3x + 1)⁴ is 12(3x + 1)³, not just 4(3x + 1)³.
• Ignoring restrictions: A quotient is defined only where its original denominator is non-zero.
• Simplifying too early: Identify the functions and apply the correct rule before expanding or cancelling.

8. Practice exercises

Attempt each question before opening the answers. Show the rule, substitution and simplification in your working.

1. Differentiate y = (x² + 3)(2x − 1).
2. Differentiate y = x² cos x.
3. Differentiate y = (x − 4)eˣ.
4. Differentiate y = x³ ln x, where x > 0.
5. Differentiate y = (2x + 5)/(x − 1).
6. Differentiate y = (x² − 1)/(x + 2).
7. Differentiate y = sin x/x.
8. Differentiate y = (4x − 3)/(x² + 1).
9. Differentiate y = x(2x + 1)³.
10. Differentiate y = (x² + 2)²/(x + 1).
11. Differentiate y = (x² + 1)eˣ.
12. Differentiate y = (3x − 2)⁴/(x² + 1).

Reveal answers and self-check

1. dy/dx = 6x² − 2x + 6.
2. dy/dx = 2x cos x − x² sin x.
3. dy/dx = eˣ(x − 3).
4. dy/dx = x²(3 ln x + 1).
5. dy/dx = −7/(x − 1)², x ≠ 1.
6. dy/dx = (x² + 4x + 1)/(x + 2)², x ≠ −2.
7. dy/dx = (x cos x − sin x)/x², x ≠ 0.
8. dy/dx = (−4x² + 6x + 4)/(x² + 1)².
9. dy/dx = (2x + 1)²(11x + 1).
10. dy/dx = (x² + 2)(−x² + 4x − 2)/(x + 1)², x ≠ −1.
11. dy/dx = eˣ(x² + 2x + 1) = eˣ(x + 1)².
12. dy/dx = (3x − 2)³(6x² + 4x + 12)/(x² + 1)².

Teacher's syllabus note

This lesson is included as an extended differentiation resource. Check the exact examination track and current syllabus before treating the product and quotient rules as compulsory UACE content: the NCDC Advanced Level Subsidiary Mathematics syllabus states that these rules should not be taught in that track, while other A-Level specifications may include them. Teachers should align the lesson with the learner's specific subject combination and examination requirements.

9. Lesson summary

• For a product, use uv′ + vu′.
• For a quotient, use (vu′ − uv′)/v².
• Use the chain rule when differentiating a composite factor.
• Keep the quotient numerator order correct and state restrictions where needed.

Coming next: Part 4 — Applications of Differentiation: Tangents, Normals, Velocity and Acceleration.

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