The Product and Quotient Rules

Differentiate two functions multiplied or divided, simplify the result into the form asked, and use it in show-that and second-derivative questions.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 38 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Two functions multiplied

Try $y = x^{2} \times x^{3}$ . That is $x^{5}$ , so $\frac{d y}{d x} = 5 x^{4}$ . Now multiply the two separate derivatives: $2 x \times 3 x^{2} = 6 x^{3}$ . That is wrong.

So you cannot just multiply the derivatives. When two functions are multiplied, each one changes while the other one waits. You add the two effects.

Formula

\frac{d}{d x} (u v) = u \frac{d v}{d x} + v \frac{d u}{d x}

$u$ — the first function, for example $x^{3}$ .
$v$ — the second function, for example $e^{2 x}$ .
$\frac{d u}{d x}, \frac{d v}{d x}$ — their separate derivatives.

Rule

Learn it. 0606 and 4037 give no calculus formulas. 4PM1 lists it with the formulas to learn.

Worked example

1
Find $\frac{d y}{d x}$ for $y = x^{3} e^{2 x}$ .
2
$u = x^{3}, \frac{d u}{d x} = 3 x^{2}$ — Name the first function and differentiate it.
3
$v = e^{2 x}, \frac{d v}{d x} = 2 e^{2 x}$ — Name the second function and differentiate it.
4
$\frac{d y}{d x} = x^{3} \cdot 2 e^{2 x} + e^{2 x} \cdot 3 x^{2}$ — u times dv/dx, plus v times du/dx.
5
$\frac{d y}{d x} = x^{2} e^{2 x} (2 x + 3)$ — Take out the common factor $x^{2} e^{2 x}$ .

Mark schemes give a method mark for two terms added in the right shape, then a mark for each correct term. So write the unsimplified line before you tidy.

Key idea

In 1675 Gottfried Leibniz checked in his notes whether the derivative of a product is just the product of the derivatives. He showed it is not, and went on to find the rule you just learned.

Products with a chain inside

Often one of the two functions needs the chain rule. Differentiate it on the side first, then put it into the product rule.

Worked example

1
$y = x 4 x + 1$ . Show that $\frac{d y}{d x} = \frac{6 x + 1}{4 x + 1}$ .
2
$u = x, \frac{d u}{d x} = 1$ — The first function.
3
$v = (4 x + 1)^{\frac{1}{2}},$ $\frac{d v}{d x} = \frac{1}{2} (4 x + 1)^{- \frac{1}{2}} \times 4 = \frac{2}{4 x + 1}$ — The chain rule on the second function.
4
$\frac{d y}{d x} = \frac{2 x}{4 x + 1} + 4 x + 1$ — u times dv/dx, plus v times du/dx.
5
$4 x + 1 = \frac{4 x + 1}{4 x + 1}$ — To add, give both terms the same bottom: a root times itself is the number inside.
6
$\frac{d y}{d x} = \frac{2 x + 4 x + 1}{4 x + 1} = \frac{6 x + 1}{4 x + 1}$ — Add the tops. This is the given answer.

Worked example

1
$y = e^{- x} cos 2 x$ . Find $\frac{d y}{d x}$ .
2
$\frac{d u}{d x} = - e^{- x}, \frac{d v}{d x} = - 2 sin 2 x$ — $u = e^{- x}$ and $v = cos 2 x$ . Watch both minus signs.
3
$\frac{d y}{d x}$ $= e^{- x} (- 2 sin 2 x)$ $+ cos 2 x (- e^{- x})$ — u times dv/dx, plus v times du/dx.
4
$\frac{d y}{d x} = - e^{- x} (cos 2 x + 2 sin 2 x)$ — Take out the common factor $- e^{- x}$ .

Common mistake

In a “show that” question, jumping to the given answer. Show the unsimplified line and the common-bottom step. The examiner must see every move.

Key idea

Differentiate each part on its own first. Then build the product rule. Then factorise or use a common bottom.

One function divided by another

When the function is a fraction with x on top and bottom, use the quotient rule. The order of the top line matters, because it has a minus sign.

Formula

\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}

$u$ — the top of the fraction.
$v$ — the bottom of the fraction. It is squared in the answer.

Rule

4PM1 gives this rule on the formula sheet, written with f(x) on top and g(x) below. 0606 and 4037 do not: learn it.

Memory trick

Bottom times d(top), minus top times d(bottom), all over bottom squared. Always start with the bottom.

Worked example

1
$y = \frac{3 x + 2}{x ^{2} + 1}$ . Find $\frac{d y}{d x}$ .
2
$u = 3 x + 2, \frac{d u}{d x} = 3;$ $v = x^{2} + 1, \frac{d v}{d x} = 2 x$ — Top and bottom, each differentiated.
3
$\frac{d y}{d x} = \frac{( x ^{2} + 1 ) ( 3 ) - ( 3 x + 2 ) ( 2 x )}{( x ^{2} + 1 ) ^{2}}$ — Bottom times d(top), minus top times d(bottom), over bottom squared.
4
$= \frac{3 x ^{2} + 3 - 6 x ^{2} - 4 x}{( x ^{2} + 1 ) ^{2}}$ — Expand the top. The minus changes both signs in the second bracket.
5
$\frac{d y}{d x} = \frac{3 - 4 x - 3 x ^{2}}{( x ^{2} + 1 ) ^{2}}$ — Collect like terms. Leave the bottom as a bracket squared.

Common mistake

Swapping the order on top: $u \frac{d v}{d x} - v \frac{d u}{d x}$ . This makes every sign wrong.

Common mistake

Forgetting the minus reaches the whole second bracket: $- (3 x + 2) (2 x) = - 6 x^{2} - 4 x$ .

Key idea

The quotient rule always starts with the bottom. Keep the bottom squared as a bracket; do not expand it.

Quotients with e, trig and ln

The same rule works with any functions you can differentiate. Factorise the top at the end: a common factor like $e^{2 x}$ makes the answer much neater.

Worked example

1
$y = \frac{e ^{2 x}}{x + 1}$ . Find $\frac{d y}{d x}$ and the gradient at $x = 0$ .
2
$\frac{d u}{d x} = 2 e^{2 x}, \frac{d v}{d x} = 1$ — $u = e^{2 x}$ on top, $v = x + 1$ below.
3
$\frac{d y}{d x} = \frac{( x + 1 ) 2 e ^{2 x} - e ^{2 x} ( 1 )}{( x + 1 ) ^{2}}$ — Bottom times d(top), minus top times d(bottom).
4
$\frac{d y}{d x} = \frac{e ^{2 x} ( 2 x + 1 )}{( x + 1 ) ^{2}}$ — Take out the common factor $e^{2 x}$ : 2x + 2 − 1 = 2x + 1.
5
$x = 0 : \frac{1 \times 1}{1} = 1$ — e to the power 0 is 1.

Rule

The next example uses ln, so it is for 0606 and 4037 only.

Worked example

1
$y = \frac{l n x}{x ^{2}}$ . Find $\frac{d y}{d x}$ in its simplest form.
2
$\frac{d y}{d x} = \frac{x ^{2} \cdot \frac{1}{x} - l n x \cdot 2 x}{x ^{4}}$ — $u = ln x$ gives $\frac{1}{x}$ ; $v = x^{2}$ gives 2x; bottom squared is $x^{4}$ .
3
$= \frac{x - 2 x l n x}{x ^{4}}$ — $x^{2} \times \frac{1}{x} = x$ .
4
$\frac{d y}{d x} = \frac{1 - 2 l n x}{x ^{3}}$ — Every term on top has a factor x. Cancel one x with the bottom.

Common mistake

Cancelling part of a sum. In $\frac{x - 2 x l n x}{x ^{4}}$ you may cancel an x only because every term on top has one. Factorise first; cancel whole factors only.

Key idea

Quotient rule, then factorise the top, then cancel whole factors. That is the form examiners ask for.

Show that: second derivatives

Some questions ask you to differentiate twice and prove a link between $y$ , $\frac{d y}{d x}$ and $\frac{d ^{2} y}{d x ^{2}}$ . Keep a common factor outside so the algebra stays short.

Worked example

1
$y = x e^{2 x}$ . Show that $\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = 0$ .
2
$\frac{d y}{d x} = x \cdot 2 e^{2 x} + e^{2 x} \cdot 1 = e^{2 x} (2 x + 1)$ — Product rule, then factorise.
3
$\frac{d ^{2} y}{d x ^{2}} = e^{2 x} \cdot 2 + (2 x + 1) \cdot 2 e^{2 x}$ — Product rule again, with $u = e^{2 x}$ and $v = 2 x + 1$ .
4
$\frac{d ^{2} y}{d x ^{2}} = e^{2 x} (4 x + 4)$ — Factorise: 2 + 4x + 2 = 4x + 4.
5
$e^{2 x} (4 x + 4) - 4 e^{2 x} (2 x + 1) + 4 x e^{2 x}$ — Substitute all three into the left-hand side.
6
$= e^{2 x} (4 x + 4 - 8 x - 4 + 4 x) = 0$ — Take out $e^{2 x}$ . The bracket collects to 0, as required.

Sometimes a quotient is easier as a product. For example, $\frac{x ^{2}}{e ^{x}} = x^{2} e^{- x}$ . Both ways give the same answer, so choose the one with less algebra.

Common mistake

Using the quotient rule when the top is a plain number. For $\frac{5}{x ^{2} + 1}$ , write $5 (x^{2} + 1)^{- 1}$ and use the chain rule. It is quicker and safer.

Key idea

In show-that questions, write each line in full and end with the exact statement you were asked to show.

Which method? Read the question

Before you differentiate, ask: what is the shape of this function?

You see	You do
Two functions of x multiplied, like $x^{3} e^{2 x}$	Product rule: $u \frac{d v}{d x} + v \frac{d u}{d x}$
x on top and bottom, like $\frac{3 x + 2}{x ^{2} + 1}$	Quotient rule: start with the bottom
A plain number over a bracket	Chain rule with a negative power
A number times one function, like $5 e^{2 x}$	No product rule: the 5 just waits in front
A product or quotient with a bracket inside	Chain rule on that part first, on the side
“Show that” or “in the form …”	Unsimplified line, then factorise or use a common bottom
A link between $y$ , $\frac{d y}{d x}$ and $\frac{d ^{2} y}{d x ^{2}}$	Differentiate twice, keep factors outside, substitute

Common mistake

Using the product rule on $5 e^{2 x}$ . A constant is not a function of x; its derivative is 0, so it just multiplies the answer.

Key idea

Product: add two terms. Quotient: subtract, starting with the bottom, over the bottom squared.

Practise the product and quotient rules now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on the product and quotient rules

Each with a full worked solution.

All 38 questions on this topic →

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.