Proof of the Quotient Rule
Dividing two differentiable functions produces a function whose derivative follows a precise pattern. The quotient rule can be derived cleanly from the product rule, illustrating how the differentiation rules form a coherent system.
Statement
Let f and g be differentiable at x with g(x) ≠ 0. Then f/g is differentiable at x and:
Proof via the Product Rule
Write the quotient as a product: let h(x) = f(x)/g(x), so f(x) = g(x)·h(x). Differentiate both sides using the product rule:
Substitute h(x) = f(x)/g(x) and solve for h′(x):
Combine over a common denominator g(x):
Divide both sides by g(x) (valid since g(x) ≠ 0):
The Condition g(x) ≠ 0
The condition g(x) ≠ 0 is not merely a technicality. If g(x₀) = 0, then f(x)/g(x) is undefined at x₀ and cannot be differentiable there. Near a zero of g, the quotient may blow up to ±∞, making the notion of a tangent line meaningless. The quotient rule applies only on the open set where g is non-zero.
Worked Example
Differentiate q(x) = (x² + 3) / (2x − 1), valid for x ≠ 1/2.
- f(x) = x² + 3, so f′(x) = 2x.
- g(x) = 2x − 1, so g′(x) = 2.
- Numerator: f′·g − f·g′ = 2x·(2x−1) − (x²+3)·2 = 4x²−2x − 2x²−6 = 2x²−2x−6.
- Denominator: [g(x)]² = (2x−1)².
Check at x = 1: q′(1) = (2−2−6)/(2−1)² = −6/1 = −6. A numerical difference quotient at x = 1 with h = 10⁻⁶ gives −6.000, confirming the result.
Frequently Asked Questions
What is the quotient rule?
If f and g are differentiable at x and g(x) ≠ 0, then (f/g)′(x) = [f′(x)·g(x) − f(x)·g′(x)] / [g(x)]². Note the subtraction order: f′g minus fg′, not the other way around.
Why must g(x) ≠ 0?
Division by zero is undefined. The quotient f/g is not defined where g(x) = 0, and neither is its derivative. The formula requires g(x) ≠ 0 in a neighbourhood of x.
Do I need to memorise the quotient rule separately from the product rule?
No. You can always write f/g as f·g⁻¹ and apply the product rule together with the power rule for g⁻¹ = (g)^{-1}: d/dx g⁻¹ = −g′/g². The quotient rule is a convenience, not a necessity.
How do I remember the correct sign in the numerator?
A common mnemonic is 'low d-high minus high d-low, square the bottom and away we go': (g·f′ − f·g′) / g². 'High' is the numerator f, 'low' is the denominator g.
Adapted and rewritten from Silas Maths source notes; formulas reviewed for CalxSolver publication.