How does the quotient rule differentiate one function divided by another?
Quick answer
For differentiable functions f and g, differentiate f/g using (f′g − g′f)/g². In words: derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared. The denominator must be nonzero where you evaluate the derivative. For example, for (x²+1)/(x−2), the rule gives (x²−4x−1)/(x−2)², valid for x ≠ 2. [S1]
The question, the details, the next step.
What is the quotient rule, and how do I remember which way round the subtraction goes?
If f and g are differentiable and g is nonzero at the point in question, then
(f/g)′ = (f′g − g′f)/g². [S1]
A useful way to say it is: “bottom times derivative of top, minus top times derivative of bottom, over bottom squared.” In symbols, that is g f′ − f g′, all over g². These two numerator versions are the same; the subtraction order matters. The denominator is the original bottom function squared, not its derivative. Also, the derivative of a quotient is generally not the quotient of the derivatives. [S1]
Before calculating, label the numerator f and denominator g. That makes it easier to keep track of which derivative belongs to which function.
Can you show every step for a small rational-function example, including where the formula is allowed?
Take h(x) = (x² + 1)/(x − 2). Set f(x) = x² + 1 and g(x) = x − 2. Their derivatives are f′(x) = 2x and g′(x) = 1.
The original denominator is zero at x = 2, so the quotient—and therefore its derivative—are not defined there. Work with x ≠ 2. [S1]
Apply the rule:
h′(x) = [f′(x)g(x) − g′(x)f(x)]/[g(x)]² = [2x(x − 2) − 1(x² + 1)]/(x − 2)².
Now expand and combine like terms:
2x(x − 2) − (x² + 1) = 2x² − 4x − x² − 1 = x² − 4x − 1.
So h′(x) = (x² − 4x − 1)/(x − 2)², for x ≠ 2. The condition comes from the original denominator; simplifying the derivative’s numerator does not make x = 2 an allowed input.
How can I check that I used the rule correctly, especially the minus sign and the domain?
Check the setup before simplifying: the first numerator term must be “derivative of top times bottom,” and the second must be “top times derivative of bottom.” Keep the minus sign between them, and square the entire original denominator. For this example, that setup is 2x(x − 2) − (x² + 1), over (x − 2)².
Then expand the numerator carefully, using parentheses after the minus sign. That minus applies to both terms inside x² + 1, which gives −x² − 1. Finally, state x ≠ 2, because the original quotient has denominator zero there. A quick substitution at an allowed value can help check the arithmetic: at x = 3, the derivative formula gives (9 − 12 − 1)/1 = −4.
Can you give me a reusable template for another quotient?
Use this fill-in pattern for y = f(x)/g(x):
- Identify top: f(x) = ___; bottom: g(x) = ___ .
- Differentiate each: f′(x) = ___; g′(x) = ___ .
- Substitute: y′ = [f′(x)g(x) − f(x)g′(x)]/[g(x)]².
- Expand and simplify the numerator, keeping the subtraction sign attached to the full second product.
- State the restriction g(x) ≠ 0, and exclude any input where the original denominator is zero.
The rule applies when the functions are differentiable and the denominator is nonzero at the point where you evaluate the derivative. [S1] This template organizes the calculation; it does not by itself establish differentiability at points where the functions or quotient are undefined.
A practical checklist
Label the numerator f(x) and denominator g(x).
Find f′(x) and g′(x) separately.
Write f′g − gf′ in the numerator and g² in the denominator.
Use parentheses so the minus sign applies to the whole second product.
Simplify, then state where the original denominator is nonzero.
A few more useful answers
Why is the denominator squared?
It is part of the quotient rule: the denominator of the derivative is the original denominator function squared, g(x)². [S1]
Can I use the quotient rule when the denominator equals zero?
No. The quotient and its derivative are not defined at a point where the original denominator is zero. [S1]
Is the derivative of f/g equal to f′/g′?
No. Use (f′g − g′f)/g² instead; differentiating the top and bottom separately and dividing is not the quotient rule. [S1]
Sources
References used for this guide.
- [S1] Differentiation Rules — Calculus Volume 1, Section 3.3 openstax.org · Checked