Product & Quotient Rules

Why these rules exist

The basic rules handle functions added together: differentiate term by term, done. It is tempting to hope multiplication works the same way — differentiate each factor and multiply. It does not, and one example proves it: x · x = x² has derivative 2x, but the product of the individual derivatives would be 1 · 1 = 1. Wrong answer.

Products and quotients change in a more interesting way — each factor’s change affects the whole, scaled by the size of the other factor — and the two rules in this guide capture that precisely. If two expressions that both contain the variable are multiplied, use the product rule; if one is divided by the other, use the quotient rule.

The Product Rule

(f · g)′ = f′g + fg′

In words: derivative of the first times the second, plus the first times the derivative of the second. Each factor takes a turn being differentiated while the other stays exactly as it is, and the two turns are added.

Applying it, step by step: x² · sin(x)

1. Name the factors. f = x² and g = sin(x).
2. Differentiate each separately. f′ = 2x (power rule) and g′ = cos(x).
3. Fill in the template f′g + fg′, keeping each piece in its slot:

d/dx[x² sin(x)] = 2x · sin(x) + x² · cos(x)

Writing the template out before filling it in feels slow, but it is the habit that prevents nearly every product rule error. Once the pieces are in their slots, the problem is just bookkeeping.

The Quotient Rule

(f / g)′ = (f′g − fg′) / g²

The shape mirrors the product rule with two crucial differences: the middle is a minus, and everything sits over the denominator squared. Because of the subtraction, order matters — the term beginning with the derivative of the numerator comes first. The classic memory aid: “low d-high minus high d-low, over low squared.”

Applying it, step by step: x / (x + 1)

1. Name the parts. Numerator f = x, denominator g = x + 1.
2. Differentiate each. f′ = 1 and g′ = 1.
3. Fill in the template (f′g − fg′) / g²:

d/dx[x/(x+1)] = [1·(x+1) − x·1] / (x+1)² = 1 / (x+1)²

Notice the numerator collapsed from (x + 1) − x to just 1. Always simplify after applying the rule — quotient rule answers usually clean up considerably.

When you don’t need either rule

A surprising number of product- and quotient-looking problems are easier after a small rewrite:

• One piece is a constant. 5x³ needs only the constant multiple rule (15x²), and x²/7 is just (1/7)x². No product or quotient rule required.
• It simplifies algebraically. x⁵/x² is x³ — one power rule step instead of a full quotient rule grind.
• The numerator is 1. 1/g(x) can be rewritten as g(x)⁻¹ and handled with the power and chain rules, which is often faster and less error-prone.

Common mistakes to avoid

• Multiplying the derivatives. Writing d/dx[x² sin x] = 2x · cos x — the tempting wrong rule from the start of this guide. How to avoid it: the moment you see two variable expressions multiplied, write the f′g + fg′ template down before doing any differentiating.
• Flipping the quotient rule’s numerator. Writing (fg′ − f′g)/g² instead of (f′g − fg′)/g². Because of the subtraction, this silently negates the whole answer. How to avoid it: say the mnemonic as you write — “low d-high minus high d-low” — and double-check that the first term starts with the numerator’s derivative.
• Reaching for the heavy rule when a constant is involved. Using the full product rule on 5x³ (it works, but invites errors) or the quotient rule on x²/7. How to avoid it: spend five seconds checking whether one piece is constant or the expression simplifies before committing to the longer rule.

Worked example, start to finish

Differentiate h(x) = ln(x) / x.

Check your work

Work these by hand, then verify each step with the derivative calculator — it names the rule at every line, so a disagreement points you to the exact step to review: