Calculus

The Product and Quotient Rules

Two rules, one of which is derived from the other. Once you have seen where the quotient rule comes from, the minus sign stops being something you have to remember.

By the Gauth editorial teamHow this content is produced and checkedUpdated 10 September 2026

The idea

Differentiation distributes over addition, so it is tempting to assume it distributes over multiplication too. It does not, and the smallest possible example proves it. Let $f (x) = x$ and $g (x) = x$ . Their product is $x^{2}$ , whose derivative is $2 x$ . But $f^{'} g^{'} = 1 \cdot 1 = 1$ . Those two functions are not the same function.

Here is why. Picture a rectangle whose width is $f$ and whose height is $g$ , both growing over time. Its area is the product. Over a short interval the width grows by $Δ f$ and the height by $Δ g$ , and the new area gains three strips: a horizontal one of area $f Δ g$ , a vertical one of area $g Δ f$ , and a tiny corner square of area $Δ f Δ g$ . Divide everything by the elapsed time and shrink the interval. The corner square is a product of two small quantities, so it vanishes faster than the others and contributes nothing in the limit. What survives is the two strips:

(f g)^{'} = f^{'} g + f g^{'}

The quotient rule is not a separate fact. It is this same rule, rearranged. Suppose $q = f / g$ , so that $f = q g$ . Differentiate that with the product rule:

f^{'} = q^{'} g + q g^{'}

Now solve for $q^{'}$ and substitute $q = f / g$ back in:

q^{'} = \frac{f ^{'} - q g ^{'}}{g} = \frac{f ^{'} - \frac{f}{g} g ^{'}}{g} = \frac{f ^{'} g - f g ^{'}}{g ^{2}}

That derivation explains both of the features people find hard to recall. The minus sign is there because the term $q g^{'}$ had to move across the equals sign. The $g^{2}$ is there because one factor of $g$ came from dividing through and the other from clearing the fraction inside. If you can reproduce those four lines you never need the mnemonic.

When you need it

Ask what the last operation is. If it is a multiplication of two expressions that both contain the variable, you need the product rule. If it is a division where the variable appears in the denominator, you need the quotient rule — or a rewrite. If the variable appears in only one of the two pieces, you probably need neither.

Product rule territory: $x^{3} sin x$ , $x e^{x}$ , $(2 x + 1) ln x$ , $x^{2} (x + 1)^{3}$ . In the last one the second factor also needs the chain rule, but the product rule still opens the problem because multiplication is the outermost operation.

Quotient rule territory: $\frac{2 x + 1}{x ^{2} + 3}$ , $\frac{e ^{2 x}}{x ^{2} + 1}$ , $\frac{sin x}{cos x}$ . In each case the denominator genuinely depends on $x$ and does not factor away.

Neither, despite appearances. A constant multiple is not a product worth the rule: $5 sin x$ differentiates to $5 cos x$ directly. A fraction with a constant denominator is a constant multiple in disguise: $\frac{x ^{3} + 2 x}{5}$ has derivative $\frac{3 x ^{2} + 2}{5}$ . And a fraction that splits should be split — $\frac{x ^{3} + 5 x}{x}$ is just $x^{2} + 5$ , so the derivative is $2 x$ . The quotient rule gets the same answer there, but through four extra lines of algebra where marks go missing.

One more rewrite worth knowing: when the numerator is a constant, use a negative power instead. $\frac{3}{x ^{2} + 1}$ becomes $3 (x^{2} + 1)^{- 1}$ , and the chain rule gives $- 6 x (x^{2} + 1)^{- 2}$ in one line. If negative and fractional exponents are not automatic for you yet, the exponent and radical rules guide covers the pattern they all follow.

The method

1. Try to avoid the rule. Spend five seconds checking for a constant denominator, a common factor, or a fraction that splits into terms. This is not laziness; it is the step that removes the most opportunities for error.

2. Label four quantities before you assemble anything. Write a short column: $f$ , $f^{'}$ , $g$ , $g^{'}$ . People who build the formula in one pass are the people who lose a term. People who fill in a template from four labelled pieces are not.

3. Differentiate each piece completely, using the chain rule where a piece is composite. Do this before substituting, so that the substitution step is pure copying.

4. Apply the formula, respecting the order.

(f g)^{'} = f^{'} g + f g^{'} (\frac{f}{g})^{'} = \frac{f ^{'} g - f g ^{'}}{g ^{2}}

In the product rule the two terms can be written in either order. In the quotient rule they cannot, because subtraction is not commutative. If you are ever unsure, test the formula on $f = x^{2}$ , $g = x$ , where the quotient is $x$ and the derivative has to come out as 1.

5. Expand the numerator, leave the denominator alone. Collect like terms on top and factor what you can. Never expand $g^{2}$ — it is already in the most useful form, and the factored version is what you need when the next part of the question asks where the derivative is zero.

Worked examples

Worked example

y = x^{3} sin x

1
Label the two factors and their derivatives before touching the formula.
$f = x^{3}, f^{'} = 3 x^{2}, g = sin x, g^{'} = cos x$
2
Substitute into f'g + fg'. Each term differentiates exactly one factor and copies the other.
$y^{'} = 3 x^{2} sin x + x^{3} cos x$
3
Factor out the lowest power of x. This is the form a marker expects, and it makes the stationary points readable.
$y^{'} = x^{2} (3 sin x + x cos x)$

Answer

y^{'} = x^{2} (3 sin x + x cos x)

Worked example

y = \frac{2 x + 1}{x ^{2} + 3}

1
The denominator depends on x and nothing cancels, so the quotient rule is the right tool. Label the four pieces.
$f = 2 x + 1, f^{'} = 2, g = x^{2} + 3, g^{'} = 2 x$
2
Assemble the numerator as f'g minus fg', in that order, over g squared.
$y^{'} = \frac{2 ( x ^{2} + 3 ) - ( 2 x + 1 ) ( 2 x )}{( x ^{2} + 3 ) ^{2}}$
3
Expand the numerator only. The denominator stays factored.
$2 x^{2} + 6 - (4 x^{2} + 2 x) = - 2 x^{2} - 2 x + 6$
4
Take out the common factor of -2 so the roots of the numerator are visible.
$y^{'} = \frac{- 2 ( x ^{2} + x - 3 )}{( x ^{2} + 3 ) ^{2}}$

Answer

y^{'} = \frac{- 2 ( x ^{2} + x - 3 )}{( x ^{2} + 3 ) ^{2}}

Worked example

y = \frac{e ^{2 x}}{x ^{2} + 1}

1
The numerator is composite, so its derivative needs the chain rule: outer is the exponential, inner is 2x.
$f = e^{2 x}, f^{'} = 2 e^{2 x}, g = x^{2} + 1, g^{'} = 2 x$
2
Substitute into the quotient rule with the pieces already differentiated.
$y^{'} = \frac{2 e ^{2 x} ( x ^{2} + 1 ) - e ^{2 x} ( 2 x )}{( x ^{2} + 1 ) ^{2}}$
3
Both numerator terms share a factor of 2e^{2x}, so take it out rather than expanding.
$y^{'} = \frac{2 e ^{2 x} [ ( x ^{2} + 1 ) - x ]}{( x ^{2} + 1 ) ^{2}}$
4
Tidy the bracket. Its discriminant is negative, so it never vanishes and the derivative is positive everywhere — the function is always increasing.
$y^{'} = \frac{2 e ^{2 x} ( x ^{2} - x + 1 )}{( x ^{2} + 1 ) ^{2}}$

Answer

y^{'} = \frac{2 e ^{2 x} ( x ^{2} - x + 1 )}{( x ^{2} + 1 ) ^{2}}

Try one of your own

Type a product or a quotient and you will get the labelled pieces and the assembled formula, not just the final expression. For functions that stack all three rules at once, the derivative calculator with steps names the rule used at every layer, and the calculus solver will take a photograph of the whole question if the expression is awkward to type.

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The classic quotient rule slip

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Where people go wrong

Assuming derivatives distribute over multiplication

Writing $(f g)^{'} = f^{'} g^{'}$ . The counterexample takes one line: with $f = g = x$ the product is $x^{2}$ with derivative $2 x$ , while $f^{'} g^{'} = 1$ . Keep that example in your head. Any time a product rule feels like too much work, it is there to remind you that the shortcut is not merely inelegant, it is wrong.

Dividing the derivatives

Writing $(f / g)^{'} = f^{'} / g^{'}$ . Same test, same verdict: $f = x^{2}$ over $g = x$ is $x$ , whose derivative is 1, but $f^{'} / g^{'} = 2 x$ . This error often surfaces because students half-remember L'Hopital's rule, which does divide derivatives — but that rule evaluates a limit of an indeterminate form, and it is not a differentiation rule at all.

Reversing the numerator, or dropping the square

The two failure modes of the quotient rule formula itself. A reversed numerator flips the sign of the entire answer, which is silent and costly — the expression still looks plausible. A missing square on the denominator changes the shape of the answer completely. Both are caught by the same ten-second test on $x^{2} / x$ , where the answer has to be exactly 1.

Using the rule where a rewrite is faster

Reaching for the quotient rule on $\frac{x ^{4} - 3 x ^{2}}{x ^{2}}$ instead of noticing it is $x^{2} - 3$ , with derivative $2 x$ . The rule is not wrong here, only slow, and every extra line is another chance to drop a sign. Scanning for a simplification first is one of the habits that separates fast, accurate work from the kind described in the twelve algebra mistakes that quietly cost marks.

Practice

Two products and two quotients. The last one derives a standard result you have probably been asked to memorise.

Product and quotient practice

Answers are checked here — nothing is sent anywhere.

1
$Differentiate y = x^{2} cos x$
2
$Differentiate y = \frac{x}{x + 1}$
3
$Differentiate y = x e^{x} and factor$
4
$Use the quotient rule on y = \frac{s i n x}{c o s x} and simplify$

Frequently asked questions

Why is the derivative of a product not the product of the derivatives?

Try it on the simplest possible case. Take f = x and g = x. Their product is x^2, whose derivative is 2x. But f' times g' is 1 times 1, which is 1. The two answers disagree everywhere except x = 0.5, so the naive rule is simply false. Changing a product changes it through both factors at once, and the product rule adds up those two contributions.

How do I remember the order in the quotient rule numerator?

Test it on a case you already know. Take f = x^2 and g = x, so the quotient is x and the derivative must be 1. The correct order gives (2x·x − x²·1)/x² = x²/x² = 1. The reversed order gives −1. Ten seconds of checking beats memorising a mnemonic you might recall backwards.

When should I rewrite instead of using the quotient rule?

Whenever the denominator is a constant, a single power of x, or the whole fraction splits into separate terms. (x^3 + 5x)/x is just x^2 + 5, and 3/(x^2+1) is easier as 3(x^2+1)^(−1) with the chain rule. The quotient rule gives the same answer but with more algebra and a squared denominator to simplify afterwards.

Does the product rule extend to three factors?

Yes. For fgh the derivative is f'gh + fg'h + fgh' — differentiate one factor at a time and leave the others alone. The pattern continues for any number of factors: you get one term per factor, and in each term exactly one factor carries a prime.

Do I have to simplify the answer?

For a quotient rule answer, expand and collect the numerator but leave the denominator as g squared, factored. Markers award the method marks for a correctly assembled expression, and an unexpanded numerator often costs the final accuracy mark. Factoring the numerator afterwards also makes the zeros visible, which is what you need if the next part of the question asks for stationary points.

Keep going

Derivative calculator with stepsEnter a product or quotient and see the labelled pieces before the formula is assembled.The chain ruleThe rule that fires inside these two whenever a factor is composite.Exponent and radical rulesThe rewrites that let you skip the quotient rule entirely.Calculus solverFull working for derivatives, integrals, and limits from a typed or photographed question.