Short on time? This page is the recap. For the derivation, the graph and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: fractions are written on one line. [A] / [B]² means all of A divided by the square of B.
Recap
- If q(x) = f(x)/g(x), with f and g differentiable and g(x) ≠ 0, then q′(x) = [g(x)f′(x) − f(x)g′(x)] / [g(x)]².
- Say it as: bottom times derivative of top, minus top times derivative of bottom, over bottom squared.
- The rule comes from the product rule applied to f = q · g, then solving for q′.
- (f/g)′ is not f′/g′. Test: x²/x = x has derivative 1, but f′/g′ = 2x.
- The derivative does not exist where the bottom is 0.
Key relationships
| Situation | Derivative |
|---|---|
| General quotient f/g | [g f′ − f g′] / g² |
| Reciprocal 1/g | −g′ / g² |
| Constant bottom, f(x)/c | f′(x)/c |
| Constant top over a power, c/xⁿ | rewrite as c x⁻ⁿ: −n c x⁻ⁿ⁻¹ |
| Values from a table at x = a | [g(a)f′(a) − f(a)g′(a)] / [g(a)]² |
| Tangent line at x = a | y = q(a) + q′(a)(x − a) |
Assumptions behind the rule
- f and g are both differentiable at the point.
- g is not 0 at the point. If g(a) = 0, the quotient is undefined at a.
- The rule is one valid method. If the fraction simplifies, or one part is a constant, rewriting first gives the same answer with less algebra.
Mistakes to avoid
- Wrong order in the numerator. This flips the sign of the whole answer.
- Differentiating the denominator instead of squaring it.
- Dropping brackets, so the minus sign only hits the first term of f g′.
- Expanding (g(x))². Leave it factored.
- Using f′/g′.
- Cancelling terms in the original fraction (only whole factors cancel).
Quick self-check
- Differentiate x/(x + 1). (1/(x + 1)²: the top of the rule is (x + 1)(1) − x(1) = 1)
- Differentiate 3/x² without the quotient rule. (Rewrite as 3x⁻², so −6x⁻³ = −6/x³)
- f(1) = 2, f′(1) = 3, g(1) = 1, g′(1) = 4. Find (f/g)′(1). ((1)(3) − (2)(4))/1² = −5)
Next: practice questions.