Sum, Product and Quotient Rules
The sum rule (f + g)′ = f′ + g′, the product rule (fg)′ = f′g + fg′ and the quotient rule (f/g)′ = (f′g − fg′)/g2 let you differentiate combinations of functions from the derivatives of the parts. In brief: (x2 sin x)′ = 2x sin x + x2 cos x and (x/(x + 1))′ = [1 · (x + 1) − x · 1]/(x + 1)2 = 1/(x + 1)2. The derivative of a product is not the product of the derivatives.
Most functions in applications are built from simple pieces by adding, multiplying and dividing. The three rules reduce the derivative of the whole to the derivatives of the pieces. This guide states the rules, proves the product rule, shows a systematic way to use them with worked examples, explains when it is better to simplify first and gives a table of results, with the typical applications.
What is the sum rule?
(f ± g)′ = f′ ± g′ and (cf)′ = cf′. The derivative of a sum is the sum of the derivatives. This follows from the limit laws: the difference quotient of f + g is the sum of the difference quotients.
What is the product rule?
(fg)′ = f′g + fg′. In words: derivative of the first times the second, plus the first times the derivative of the second.
Proof: [f(x + h)g(x + h) − f(x)g(x)]/h = [f(x + h) − f(x)]/h · g(x + h) + f(x) · [g(x + h) − g(x)]/h. As h → 0, g(x + h) → g(x) (differentiable functions are continuous), so the limit is f′g + fg′. ∎ For three factors: (fgh)′ = f′gh + fg′h + fgh′.
Example 1
(x2 sin x)′ = 2x sin x + x2 cos x. And (x ex)′ = ex + x ex = (1 + x)ex. And (x ln x)′ = ln x + 1.
Example 2: checking by expanding
((x + 1)(x2 − 3))′ = 1 · (x2 − 3) + (x + 1) · 2x = x2 − 3 + 2x2 + 2x = 3x2 + 2x − 3. Check: the product expands to x3 + x2 − 3x − 3, with derivative 3x2 + 2x − 3 ✓.
Example 3: three factors
(x · ex · cos x)′ = ex cos x + x ex cos x − x ex sin x = ex(cos x + x cos x − x sin x).
What is the quotient rule?
(f/g)′ = (f′g − fg′)/g2, where g ≠ 0. In words: (derivative of the top times the bottom, minus the top times the derivative of the bottom), all over the bottom squared. A memory aid: “low d-high minus high d-low, over low-low”.
The rule follows from the product rule applied to f · (1/g), together with (1/g)′ = −g′/g2.
Example 4
(x/(x + 1))′ = [1 · (x + 1) − x · 1]/(x + 1)2 = 1/(x + 1)2.
Example 5
(sin x/x)′ = (x cos x − sin x)/x2. And (ex/x2)′ = (ex x2 − ex · 2x)/x4 = ex(x − 2)/x3.
Example 6: tangent from the quotient rule
(tan x)′ = (sin x/cos x)′ = (cos x · cos x − sin x · (−sin x))/cos2x = 1/cos2x.
Example 7: a rational function
f(x) = (2x + 3)/(x2 + 1): f′ = [2(x2 + 1) − (2x + 3)(2x)]/(x2 + 1)2 = (2x2 + 2 − 4x2 − 6x)/(x2 + 1)2 = (−2x2 − 6x + 2)/(x2 + 1)2. At x = 0: f′(0) = 2.
How do you choose a strategy?
- Simplify first if possible: split fractions with a single-term denominator, expand products of polynomials.
- Identify the outermost operation: a sum, a product or a quotient.
- Apply the corresponding rule, and compute the derivatives of the parts.
- Simplify the result by factoring common terms.
Example 8: simplify first
(x3 + 4x)/x = x2 + 4 (for x ≠ 0), with the derivative 2x. Using the quotient rule would give the same: [(3x2 + 4)x − (x3 + 4x)]/x2 = (2x3)/x2 = 2x ✓, but with more work.
Example 9: a constant in the numerator
(3/(x2 + 1))′: the quotient rule gives [0 − 3 · 2x]/(x2 + 1)2 = −6x/(x2 + 1)2. For compositions such as 1/(x2 + 1), the chain rule gives the same.
More worked examples
Example 10: horizontal tangent
Where does f(x) = x e−x have a horizontal tangent? f′ = e−x − x e−x = e−x(1 − x) (using (e−x)′ = −e−x from the chain rule). It is 0 at x = 1, where f(1) = 1/e ≈ 0.368.
Example 11: rate of change in a model
The revenue R(p) = p · q(p) with demand q(p) = 100 − 2p: R′(p) = q + p q′ = 100 − 2p − 2p = 100 − 4p. Revenue is maximised at R′ = 0, p = 25, giving R = 25 · 50 = 1,250.
Example 12: second derivative of a product
(fg)″ = f″g + 2f′g′ + fg″. For f = x2 and g = ex: 2ex + 4x ex + x2ex = ex(x2 + 4x + 2).
Table of derivatives
| Function | Rule | Derivative |
|---|---|---|
| x3 + sin x | sum | 3x2 + cos x |
| x2 ex | product | ex(x2 + 2x) |
| x ln x | product | ln x + 1 |
| x/(x + 1) | quotient | 1/(x + 1)2 |
| sin x/x | quotient | (x cos x − sin x)/x2 |
| ex/x2 | quotient | ex(x − 2)/x3 |
Where do the rules come from?
The product rule was found independently by Newton and Leibniz. Leibniz first thought that (fg)′ = f′g′ and corrected himself in 1675, which is why the rule is sometimes called Leibniz’s rule. The general formula for the nth derivative of a product, (fg)(n) = Σ C(n, k) f(k)g(n−k), is also named after him and resembles the binomial theorem. The quotient rule was developed soon after.
Where are they used?
The product rule is used in physics when a quantity is a product of changing factors (for example momentum p = mv with changing mass, or power P = VI). In economics, revenue is a product of price and quantity, and the quotient rule gives the rate of change of average cost A(q) = C(q)/q: A′ = (q C′ − C)/q2. The average cost is minimal where the marginal cost equals the average cost.
Which mistakes are common?
Differentiating factors separately
(x · sin x)′ ≠ 1 · cos x. Use f′g + fg′.
Wrong order or sign in the quotient rule
The numerator is f′g − fg′, not fg′ − f′g. Check with a simple case: (1/x)′ = (0 · x − 1 · 1)/x2 = −1/x2 ✓.
Forgetting to square the denominator
The denominator of the result is g2.
Using the quotient rule when not needed
For a constant denominator, such as (x2 + 1)/3, use the constant multiple rule: (2x)/3.