Quotient Rule

A2 · P3 · 11 min

The quotient rule differentiates one function divided by another, such as $\dfrac{2x - 4}{3x + 2}$ , $\dfrac{\ln x}{x}$ or $\dfrac{\sin x}{2 - \cos x}$ . It is the product rule in disguise, but it comes with a minus sign and an order that matters, so it is where many marks are lost. It also proves the derivatives of $\tan x$ , $\sec x$ , $\operatorname{cosec} x$ and $\cot x$ , and appears in almost every P3 stationary-point question involving a fraction.

Where the rule comes from

Write the quotient as a product with a negative power: $\dfrac{u}{v} = u \cdot v^{-1}$ . Apply the product rule, using the chain rule on $v^{-1}$ :

\frac{d}{dx}\left(uv^{-1}\right) = u \cdot \left(-v^{-2}\frac{dv}{dx}\right) + v^{-1}\frac{du}{dx} = \frac{1}{v}\frac{du}{dx} - \frac{u}{v^2}\frac{dv}{dx}

Put both terms over $v^2$ :

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}

So you never strictly need the quotient rule. But for fractions with a function of $x$ on the bottom, it is the cleanest way to get a single fraction, which is the form you need for solving $\dfrac{dy}{dx} = 0$ .

Quotient rule

If $y = \dfrac{u}{v}$ , where $u$ and $v$ are functions of $x$ , then

\frac{dy}{dx} = \frac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}

The order matters: it starts with $v\dfrac{du}{dx}$ , the bottom times the derivative of the top. A common memory aid is "low d-high minus high d-low, over low squared".

A check with a case you already know: $\dfrac{x^3}{x} = x^2$ , derivative $2x$ . The rule gives $\dfrac{x \cdot 3x^2 - x^3 \cdot 1}{x^2} = \dfrac{2x^3}{x^2} = 2x$ . If you swap the order in the numerator you get $-2x$ , which shows why the order matters.

Setting out

Using the quotient rule

Write $u =$ (numerator), $v =$ (denominator).
Find $\dfrac{du}{dx}$ and $\dfrac{dv}{dx}$ , with the chain rule where needed.
Write $\dfrac{dy}{dx} = \dfrac{v u' - u v'}{v^2}$ , putting brackets round every piece before you expand.
Simplify the numerator only: expand, collect, factorise. Leave the denominator as $v^2$ in factorised form.
For stationary points, set the numerator equal to zero. The denominator never makes a fraction zero.

The numerator does all the work

Because $v^2$ is a square, it is positive wherever the function is defined. So the sign of $\dfrac{dy}{dx}$ is the sign of the numerator, and $\dfrac{dy}{dx} = 0$ exactly when the numerator is zero. This is why step 4 says simplify the numerator only: expanding $(3x + 2)^2$ in the denominator achieves nothing.

A neat consequence: for $\dfrac{ax + b}{cx + d}$ the numerator always simplifies to a constant, $ad - bc$ . So the curve is either increasing everywhere or decreasing everywhere on each branch, and has no stationary points.

When not to use the quotient rule

The quotient rule is reliable but long. Three situations have a faster route.

Situation	Example	Faster route
Constant numerator	$\dfrac{5}{(2x+1)^3}$	Write as $5(2x + 1)^{-3}$ , chain rule: $-\dfrac{30}{(2x+1)^4}$
Single-term denominator that divides in	$\dfrac{x^2 + 3x - 1}{x}$	Split: $x + 3 - x^{-1}$ , derivative $1 + x^{-2}$
Exponential denominator	$\dfrac{x^2}{e^x}$	Write as $x^2e^{-x}$ , product rule
Log of a quotient	$\ln\dfrac{x+1}{x-1}$	Log laws: $\ln(x+1) - \ln(x - 1)$

If the question says "use the quotient rule", use it. Otherwise choose whichever is quickest and safest for you.

Proving the trigonometric derivatives

The quotient rule gives the derivative of $\tan x$ from those of $\sin x$ and $\cos x$ :

\frac{d}{dx}\tan x = \frac{d}{dx}\frac{\sin x}{\cos x} = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x

The same idea gives $\sec x$ , $\operatorname{cosec} x$ and $\cot x$ ; see differentiating trigonometric functions. "Prove that $\dfrac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x$ " is a standard short question.

Worked examples

A linear over a linear

Find $\dfrac{dy}{dx}$ when $y = \dfrac{2x - 4}{3x + 2}$ , and explain why the curve has no stationary points.

Solution

$u = 2x - 4$ , $u' = 2$ ; $\ v = 3x + 2$ , $v' = 3$ .

\frac{dy}{dx} = \frac{(3x + 2)(2) - (2x - 4)(3)}{(3x+2)^2} = \frac{6x + 4 - 6x + 12}{(3x + 2)^2} = \frac{16}{(3x+2)^2}

The numerator is the constant $16$ , which is never zero, so there are no stationary points. In fact $\dfrac{dy}{dx} > 0$ wherever it is defined.

Exponential over a power

Find the exact coordinates of the stationary point of $y = \dfrac{e^{2x}}{x}$ , $x \ne 0$ .

Solution

$u = e^{2x}$ , $u' = 2e^{2x}$ ; $\ v = x$ , $v' = 1$ .

\frac{dy}{dx} = \frac{x \cdot 2e^{2x} - e^{2x}\cdot 1}{x^2} = \frac{e^{2x}(2x - 1)}{x^2}

Since $e^{2x} > 0$ , the numerator is zero only when $x = \tfrac{1}{2}$ . Then $y = \dfrac{e^1}{1/2} = 2e$ .

The stationary point is $\left(\tfrac{1}{2}, 2e\right)$ .

Where a function is increasing

Given $y = \dfrac{x}{x^2 + 4}$ , find the coordinates of the stationary points and the set of values of $x$ for which $y$ is increasing.

Solution

$u = x$ , $u' = 1$ ; $\ v = x^2 + 4$ , $v' = 2x$ .

\frac{dy}{dx} = \frac{(x^2 + 4) - x(2x)}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}

Stationary where $4 - x^2 = 0$ : $x = \pm 2$ , giving $\left(2, \tfrac{1}{4}\right)$ and $\left(-2, -\tfrac{1}{4}\right)$ .

The denominator is positive, so $y$ is increasing where $4 - x^2 > 0$ , that is $-2 < x < 2$ .

y = x / (x^2 + 4)

The curve $y = \dfrac{x}{x^2 + 4}$ rises between its minimum at $x = -2$ and its maximum at $x = 2$ .

A trigonometric quotient

Find the $x$ -coordinates of the stationary points of $y = \dfrac{\sin x}{2 - \cos x}$ for $0 \le x < 2\pi$ , and the exact $y$ -coordinate of each.

Solution

$u = \sin x$ , $u' = \cos x$ ; $\ v = 2 - \cos x$ , $v' = \sin x$ .

\frac{dy}{dx} = \frac{(2 - \cos x)\cos x - \sin x \cdot \sin x}{(2 - \cos x)^2} = \frac{2\cos x - (\cos^2 x + \sin^2 x)}{(2 - \cos x)^2} = \frac{2\cos x - 1}{(2 - \cos x)^2}

Zero when $\cos x = \tfrac{1}{2}$ : $x = \tfrac{\pi}{3}$ or $x = \tfrac{5\pi}{3}$ .

At $x = \tfrac{\pi}{3}$ : $y = \dfrac{\sqrt{3}/2}{3/2} = \dfrac{\sqrt{3}}{3}$ . At $x = \tfrac{5\pi}{3}$ : $y = -\dfrac{\sqrt{3}}{3}$ .

A maximum, then a normal

The curve $y = \dfrac{\ln x}{x}$ , for $x > 0$ , has one stationary point.

(a) Find its exact coordinates and show that it is a maximum.

(b) Find the equation of the normal to the curve at the point where it crosses the $x$ -axis.

Solution

(a) $u = \ln x$ , $u' = \dfrac{1}{x}$ ; $\ v = x$ , $v' = 1$ .

\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \ln x}{x^2} = \frac{1 - \ln x}{x^2}

Zero when $\ln x = 1$ , so $x = e$ and $y = \dfrac{1}{e}$ .

Differentiate again, with $u = 1 - \ln x$ , $u' = -\dfrac{1}{x}$ , $v = x^2$ , $v' = 2x$ :

\frac{d^2y}{dx^2} = \frac{x^2\left(-\frac{1}{x}\right) - (1 - \ln x)(2x)}{x^4} = \frac{-x - 2x + 2x\ln x}{x^4} = \frac{2\ln x - 3}{x^3}

At $x = e$ : $\dfrac{2 - 3}{e^3} = -\dfrac{1}{e^3} < 0$ . So $\left(e, \tfrac{1}{e}\right)$ is a maximum.

(b) $y = 0$ when $\ln x = 0$ , so at $(1, 0)$ . Gradient there: $\dfrac{1 - 0}{1} = 1$ , so the normal has gradient $-1$ . Normal: $y = -(x - 1)$ , i.e. $x + y = 1$ .

Exam-hard: finding an unknown constant

The curve $y = \dfrac{x + a}{x^2 + 3}$ , where $a$ is a constant, has a stationary point at $x = 1$ .

(a) Find $a$ .

(b) Find the coordinates of the other stationary point, and determine the nature of each.

Solution

(a) $u = x + a$ , $u' = 1$ ; $\ v = x^2 + 3$ , $v' = 2x$ .

\frac{dy}{dx} = \frac{(x^2 + 3) - (x + a)(2x)}{(x^2 + 3)^2} = \frac{3 - 2ax - x^2}{(x^2+3)^2}

At $x = 1$ the numerator is zero: $3 - 2a - 1 = 0$ , so $a = 1$ .

(b) With $a = 1$ the numerator is $3 - 2x - x^2 = -(x + 3)(x - 1)$ . The other stationary point is at $x = -3$ , where $y = \dfrac{-2}{12} = -\dfrac{1}{6}$ . At $x = 1$ , $y = \dfrac{2}{4} = \dfrac{1}{2}$ .

Nature by the sign of the numerator $-(x + 3)(x - 1)$ , a downward parabola with roots $-3$ and $1$ : negative for $x < -3$ , positive for $-3 < x < 1$ , negative for $x > 1$ .

So $\left(-3, -\tfrac{1}{6}\right)$ is a minimum (gradient $-$ to $+$ ) and $\left(1, \tfrac{1}{2}\right)$ is a maximum (gradient $+$ to $-$ ).

Common mistakes

Wrong order in the numerator. $\dfrac{u v' - v u'}{v^2}$ gives the right answer with the wrong sign. Start with the denominator: $v u'$ .
Missing brackets. $(3x + 2)(2) - (2x - 4)(3)$ : without the second bracket, $-2x - 4 \times 3$ becomes $-6x - 12$ instead of $-6x + 12$ . Bracket every piece, then expand.
Forgetting to square the denominator.
Expanding the denominator. It wastes time and hides factors that might cancel with the numerator.
Setting the denominator to zero when looking for stationary points. A fraction is zero only when its numerator is zero.
Cancelling incorrectly. In $\dfrac{x \cdot 2e^{2x} - e^{2x}}{x^2}$ you cannot cancel an $x$ from one term of the numerator only.

Exam tip

Many mark schemes award the first mark for the correct structure $\dfrac{v u' - u v'}{v^2}$ with your $u'$ and $v'$ ; the second for correct derivatives; the third for a correctly simplified result. Show the unsimplified quotient before you tidy it.
"Show that $\dfrac{dy}{dx} = \dfrac{\ldots}{\ldots}$ ": your final line must match exactly. If the printed answer has $\cos^2 x + \sin^2 x$ replaced by $1$ , show that step.
For "find the set of values for which $y$ is increasing", state that the denominator is positive and solve "numerator $> 0$ ".
Leave the denominator factorised in every answer unless told otherwise.

Summary

$\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v u' - u v'}{v^2}$ , with the denominator term first.
It is the product rule applied to $u \cdot v^{-1}$ .
Bracket each part before expanding; simplify the numerator, leave $v^2$ alone.
$\dfrac{dy}{dx} = 0 \iff$ numerator $= 0$ , and the sign of $\dfrac{dy}{dx}$ is the sign of the numerator.
Avoid the rule for constant numerators, single-term denominators, exponential denominators and logs of quotients.
The quotient rule proves $\dfrac{d}{dx}\tan x = \sec^2 x$ and the other reciprocal trig derivatives.

Practice

Question

Differentiate $\dfrac{3x + 1}{x - 2}$ .
Differentiate $\dfrac{x^2}{2x + 1}$ , simplifying the numerator fully.
Show that $\dfrac{d}{dx}\left(\dfrac{e^x}{1 + e^x}\right) = \dfrac{e^x}{(1 + e^x)^2}$ .
Differentiate $\dfrac{1 + \ln x}{x}$ .
Show that $\dfrac{d}{dx}\left(\dfrac{\cos x}{1 + \sin x}\right) = -\dfrac{1}{1 + \sin x}$ .
Prove that $\dfrac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x$ .
Find the coordinates of the stationary points of $y = \dfrac{x^2}{x - 1}$ and determine their nature.
The curve $y = \dfrac{2x - 1}{x^2 + 2}$ has two stationary points. Find their coordinates and the set of values of $x$ for which $y$ is increasing.
The curve $y = \dfrac{e^{2x}}{x^2 + k}$ , where $k$ is a positive constant, is defined for all $x$ . Find the set of values of $k$ for which the curve has two stationary points.

Answers

$\dfrac{(x - 2)(3) - (3x + 1)(1)}{(x - 2)^2} = \dfrac{-7}{(x - 2)^2}$ .
$\dfrac{(2x + 1)(2x) - x^2(2)}{(2x+1)^2} = \dfrac{2x^2 + 2x}{(2x+1)^2} = \dfrac{2x(x+1)}{(2x+1)^2}$ .
$u = e^x$ , $v = 1 + e^x$ , $u' = v' = e^x$ : $\dfrac{(1 + e^x)e^x - e^x \cdot e^x}{(1 + e^x)^2} = \dfrac{e^x + e^{2x} - e^{2x}}{(1 + e^x)^2} = \dfrac{e^x}{(1+e^x)^2}$ .
$\dfrac{x\cdot\frac{1}{x} - (1 + \ln x)\cdot 1}{x^2} = \dfrac{1 - 1 - \ln x}{x^2} = -\dfrac{\ln x}{x^2}$ .
$\dfrac{(1 + \sin x)(-\sin x) - \cos x\cos x}{(1 + \sin x)^2} = \dfrac{-\sin x - (\sin^2 x + \cos^2 x)}{(1 + \sin x)^2} = \dfrac{-(1 + \sin x)}{(1 + \sin x)^2} = -\dfrac{1}{1 + \sin x}$ .
$\cot x = \dfrac{\cos x}{\sin x}$ , so $\dfrac{d}{dx}\cot x = \dfrac{\sin x(-\sin x) - \cos x\cos x}{\sin^2 x} = \dfrac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = -\dfrac{1}{\sin^2 x} = -\operatorname{cosec}^2 x$ .
$\dfrac{dy}{dx} = \dfrac{(x - 1)(2x) - x^2}{(x-1)^2} = \dfrac{x(x - 2)}{(x - 1)^2}$ . Stationary points $(0, 0)$ and $(2, 4)$ . The numerator $x(x-2)$ is positive for $x < 0$ , negative for $0 < x < 2$ (excluding $x = 1$ ), positive for $x > 2$ . So $(0, 0)$ is a maximum and $(2, 4)$ is a minimum.
$\dfrac{dy}{dx} = \dfrac{2(x^2 + 2) - (2x - 1)(2x)}{(x^2 + 2)^2} = \dfrac{-2x^2 + 2x + 4}{(x^2+2)^2} = \dfrac{-2(x - 2)(x + 1)}{(x^2 + 2)^2}$ . Stationary points $\left(2, \tfrac{1}{2}\right)$ and $(-1, -1)$ . Increasing where $-2(x-2)(x+1) > 0$ , i.e. $(x - 2)(x + 1) < 0$ : $-1 < x < 2$ .
$\dfrac{dy}{dx} = \dfrac{(x^2 + k)2e^{2x} - e^{2x}(2x)}{(x^2 + k)^2} = \dfrac{2e^{2x}(x^2 - x + k)}{(x^2 + k)^2}$ . Since $e^{2x} > 0$ , stationary points satisfy $x^2 - x + k = 0$ . Two distinct roots need discriminant $1 - 4k > 0$ , so $k < \tfrac{1}{4}$ . With $k > 0$ given: $0 < k < \tfrac{1}{4}$ .

Where the rule comes from

Setting out

The numerator does all the work

When not to use the quotient rule

Proving the trigonometric derivatives

Worked examples

Practice

Console