Practice Maths

The Quotient Rule and Combined Rules

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Key Terms

The quotient rule applies when a function is the ratio of two differentiable functions: y = f(x)/g(x).
Formula: d/dx[f/g] = (f′g − fg′) / g². “Low d-high minus high d-low, all over low squared.”
The order of subtraction matters — it is f′g − fg′, not fg′ − f′g.
Many quotients can also be differentiated by rewriting as a product with a negative power (using the chain rule), but the quotient rule is often more direct.
Combined rules
means applying chain, product, and quotient rules together for complex functions.
📚 QCAA Formula
Quotient rule: d/dx[f(x)/g(x)] = [f′(x)g(x) − f(x)g′(x)] / [g(x)]²
y = f/g f′, g′ dy/dx = (f′g − fg′)/g²
(x + 1)/(x − 2)f′ = 1, g′ = 1[1(x − 2) − (x + 1)(1)] / (x − 2)² = −3/(x − 2)²
x²/(3x + 1)f′ = 2x, g′ = 3[2x(3x + 1) − 3x²] / (3x + 1)² = (3x² + 2x)/(3x + 1)²
1/(x² + 1)f′ = 0, g′ = 2x[0 · (x² + 1) − 1 · 2x]/(x² + 1)² = −2x/(x² + 1)²
y = f(x) / g(x) quotient → dy/dx = [ f'(x) · g(x) − f(x) · g'(x) ] / [g(x)]² f'g ← differentiate numerator fg' ← differentiate denominator
Hot Tip The memory phrase “lo d-hi minus hi d-lo, over lo squared” (where lo = denominator, hi = numerator) reliably gives the correct sign. Write out f, g, f′, g′ before substituting — this avoids sign errors.

Worked Example 1 — Simple rational function

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

Identify: f = x + 1, f′ = 1; g = x − 2, g′ = 1.

Apply: dy/dx = [1(x − 2) − (x + 1)(1)] / (x − 2)² = (x − 2 − x − 1)/(x − 2)²

Answer: dy/dx = −3/(x − 2)²

Worked Example 2 — Polynomial over linear

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

Identify: f = x², f′ = 2x; g = 3x + 1, g′ = 3.

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

Answer: dy/dx = (3x² + 2x)/(3x + 1)²

Worked Example 3 — Chain rule inside quotient rule

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

Identify: f = (x² + 1)³, f′ = 3(x² + 1)²(2x) = 6x(x² + 1)²; g = 2x − 1, g′ = 2.

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

Factor: = (x² + 1)²[6x(2x − 1) − 2(x² + 1)] / (2x − 1)²

= (x² + 1)²[12x² − 6x − 2x² − 2] / (2x − 1)² = (x² + 1)²(10x² − 6x − 2) / (2x − 1)²

Motivation: Why the Quotient Rule?

When a function is written as a fraction of two expressions in x, neither factor is constant, so you cannot differentiate numerator and denominator separately. The quotient rule provides the correct formula derived from first principles by expressing f/g as f × g−1 and applying the product rule.

Deriving the Quotient Rule

Write y = f/g = f × g−1. By the product rule: dy/dx = f′ × g−1 + f × (−g−2g′) = f′/g − fg′/g² = (f′g − fg′)/g². This is the quotient rule.

Step-by-Step Method

  1. Label: Write f = numerator, g = denominator.
  2. Differentiate: Find f′ and g′ separately (use chain rule if needed).
  3. Substitute: dy/dx = (f′g − fg′) / g².
  4. Simplify: Expand the numerator, collect like terms, and factor if possible.
Worked Example 1 — Polynomial divided by polynomial

Differentiate y = (2x − 3)/x².

f = 2x − 3, f′ = 2. g = x², g′ = 2x.

dy/dx = [2 × x² − (2x − 3) × 2x] / x4

= [2x² − 4x² + 6x] / x4 = (−2x² + 6x)/x4

= x(−2x + 6)/x4 = (−2x + 6)/x³ = 2(3 − x)/x³

Worked Example 2 — Turning point using quotient rule

Find the x-coordinate of any stationary point of y = x/(x² + 1).

f = x, f′ = 1; g = x² + 1, g′ = 2x.

dy/dx = [1(x² + 1) − x(2x)] / (x² + 1)² = (x² + 1 − 2x²)/(x² + 1)² = (1 − x²)/(x² + 1)²

Set dy/dx = 0: 1 − x² = 0 → x² = 1 → x = ±1.

Combined Rules: Chain + Product + Quotient

Complex functions may require multiple rules applied together. The general approach:

  • If the function is a quotient, use the quotient rule as the outer operation.
  • If the numerator or denominator is itself a composite function, apply the chain rule when differentiating that part.
  • If the numerator or denominator is a product, apply the product rule to differentiate it.

Always work systematically: identify all components, differentiate each one, then assemble using the appropriate formula.

Worked Example 3 — Combined quotient and product rule

Differentiate y = x(x + 1) / (x² + 1).

Numerator f = x(x + 1) = x² + x, so f′ = 2x + 1. g = x² + 1, g′ = 2x.

dy/dx = [(2x + 1)(x² + 1) − (x² + x)(2x)] / (x² + 1)²

Numerator: 2x³ + 2x + x² + 1 − 2x³ − 2x² = −x² + 2x + 1

dy/dx = (−x² + 2x + 1) / (x² + 1)²

💡 Key Reminder: The most common error with the quotient rule is reversing the subtraction in the numerator, writing fg′ − f′g instead of f′g − fg′. Always write out all four components (f, g, f′, g′) before substituting, and double-check the sign.

Summary

Quotient rule: d/dx[f/g] = (f′g − fg′)/g². Steps: label f, g; find f′, g′ (chain rule if needed); substitute; simplify. Combined rules: apply quotient rule as the outer structure, using chain and product rules within the numerator and denominator as required.

Mastery Practice

See Answers ➔
  1. Fluency

    Differentiate each function using the quotient rule.

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

    Differentiate each rational function. Simplify your answer fully.

    1. (a) y = (3x + 2)/(x + 1)
    2. (b) y = x/(x² + 4)
    3. (c) y = (x² − 1)/(x + 1)   [Hint: simplify by factoring the numerator first]
  3. Fluency

    Differentiate. Express your answers with positive indices where relevant.

    1. (a) y = 4/x²
    2. (b) y = (x + 1)/x³
    3. (c) y = 3/√x
  4. Understanding

    Find all x-values where dy/dx = 0 (turning points) for each function.

    Method: Apply the quotient rule, then set the numerator of dy/dx equal to zero.
    1. (a) y = x/(x² + 1)
    2. (b) y = x²/(x − 1)
    3. (c) y = (x − 1)/(x² + 3)
  5. Understanding

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

    1. (a) y = (x + 1)/(x − 2) at x = 3
    2. (b) y = x²/(x + 1) at x = 1
    3. (c) y = 1/(x² + 1) at x = 0
  6. Understanding

    Simplify the derivative. Show that the quotient rule result can be simplified to the given form.

    1. (a) Show that d/dx[(x² + 1)/(x)] = (x² − 1)/x². Verify by rewriting y = x + x−1 first.
    2. (b) Show that d/dx[(2x − 3)/x²] = 2(3 − x)/x³.
  7. Understanding

    Chain rule combined with quotient rule.

    Differentiate each function (the numerator or denominator requires chain rule).

    1. (a) y = (2x + 1)³ / x
    2. (b) y = √x / (x² + 1)
    3. (c) y = (x² + 3)² / (x − 1)
  8. Problem Solving

    Product rule combined with quotient rule.

    Differentiate each function, which involves both a product structure and a quotient.

    1. (a) y = x(x + 1) / (x² + 1)
    2. (b) y = x²(x − 1) / (2x + 3)
  9. Problem Solving

    Related rates using the quotient rule.

    Challenge. The concentration of a drug in the bloodstream at time t hours is modelled by C(t) = 5t / (t² + 4) (in mg/L).
    1. (a) Find C′(t) using the quotient rule.
    2. (b) Find the time at which the concentration is at its maximum (set C′(t) = 0).
    3. (c) Find the maximum concentration.
    4. (d) Is the concentration increasing or decreasing at t = 3? Justify your answer.
  10. Problem Solving

    Full combined rules question.

    For the function f(x) = x²(x + 1)3 / (2x − 1):

    1. (a) Identify the numerator u and denominator v. Find u′ using the product and chain rule.
    2. (b) Apply the quotient rule to find f′(x).
    3. (c) Factor f′(x) as fully as possible.
    4. (d) Find the gradient of the curve at x = 1.