Quotient rule | Year 12 | Ascent Library

01 / LEARN

Understand the idea.

A ratio has two changing parts

For a quotient $y = \frac{u}{v}$ with $v \neq 0$ , both numerator and denominator may change. Its derivative is $\frac{u^{'} v - u v^{'}}{v^{2}}$ . Write the subtraction in this order and bracket the entire numerator.

Record excluded inputs before differentiating. The derivative describes gradients of the original curve; a simplified expression cannot supply a gradient where that curve has no point.

Simplification and checking

First check whether division gives simpler powers. For $x \neq 0$ , $\frac{x^{2} + 1}{x} = x + x^{- 1}$ , so termwise differentiation avoids a quotient calculation. Retain the same excluded input after simplifying.

For the supplied exponential and trigonometric factors, use ${(e^{x})}^{'} = e^{x}$ , ${(sin x)}^{'} = cos x$ and ${(cos x)}^{'} = - sin x$ in radians. Expand or factor to compare equivalent derivative answers.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A rational derivative with its domain

Differentiate $y = (x^{2} + 1) / (x - 1)$ .

Worked solution

1
Record the domain and factor derivatives.
$x \neq 1, u = x^{2} + 1, u^{'} = 2 x, v = x - 1, v^{'} = 1$
The original denominator must be nonzero.
2
Apply the quotient rule.
$y^{'} = \frac{2 x (x - 1) - (x^{2} + 1) \cdot 1}{{(x - 1)}^{2}}$
The numerator is top derivative times bottom, minus top times bottom derivative.
3
Simplify the whole numerator.
$\begin{aligned} 2 x (x - 1) - (x^{2} + 1) & = 2 x^{2} - 2 x - x^{2} - 1 \\ = x^{2} - 2 x - 1 \end{aligned}$
The subtraction changes the signs of both terms in the bracket.
4
State the derivative with its restriction.
$y^{'} = \frac{x^{2} - 2 x - 1}{{(x - 1)}^{2}}, x \neq 1$
The excluded input remains excluded.

Answer

$\frac{x^{2} - 2 x - 1}{{(x - 1)}^{2}}$ , $x \neq 1$

EXAMPLE 02

A tangent to a quotient

Find the tangent to $y = x / (x + 2)$ at $x = 0$ .

Worked solution

1
Find the contact point.
$y (0) = \frac{0}{0 + 2} = 0$
The original function supplies the point (0,0).
2
Differentiate the quotient.
$\begin{aligned} y^{'} & = \frac{1 (x + 2) - x (1)}{{(x + 2)}^{2}} \\ = \frac{2}{{(x + 2)}^{2}}, x \neq - 2 \end{aligned}$
The two x terms in the numerator cancel.
3
Find the gradient at zero.
$y^{'} (0) = \frac{2}{2^{2}} = \frac{1}{2}$
The stated input lies in the curve domain.
4
Write the tangent.
$y - 0 = \frac{1}{2} (x - 0) ⟹ y = \frac{x}{2}$
A line through the origin with gradient one half has this equation.

Answer

$y = x / 2$

03 / YOUR TURN

Try it yourself.

Check your understanding

Differentiate $1 / (x + 3)$ .

Show answer & reasoning

$- 1 / {(x + 3)}^{2}$ , $x \neq - 3$

Here $u = 1$ has derivative $0$ , while $v = x + 3$ has derivative $1$ . Thus the numerator is $0 (x + 3) - 1 = - 1$ , giving $- \frac{1}{{(x + 3)}^{2}}$ for $x \neq - 3$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Differentiate $\frac{3 x + 1}{x + 2}$ .
2 marks
Differentiate $\frac{x^{2} - 4}{x}$ efficiently.
2 marks
Find $f^{'} (2)$ if $f (x) = \frac{x^{2}}{x + 1}$ .
2 marks
Differentiate $\frac{e^{x}}{x + 1}$ .
2 marks
Find the tangent to $y = \frac{x + 1}{x - 1}$ at $x = 0$ .
3 marks
Why is the derivative formula for $\frac{x^{2} - 1}{x - 1}$ not valid at $x = 1$ , even though cancellation gives $x + 1$ elsewhere?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Differentiate $\frac{3 x + 1}{x + 2}$ .
2 marks
The numerator is $3 (x + 2) - (3 x + 1) = 5$ .
Divide by ${(x + 2)}^{2}$ .
$\frac{5}{{(x + 2)}^{2}}$ , $x \neq - 2$
Differentiate $\frac{x^{2} - 4}{x}$ efficiently.
2 marks
Rewrite as $x - 4 x^{- 1}$ .
Use the power rule.
$1 + \frac{4}{x^{2}}$ , $x \neq 0$
Find $f^{'} (2)$ if $f (x) = \frac{x^{2}}{x + 1}$ .
2 marks
The derivative is $\frac{2 x (x + 1) - x^{2}}{{(x + 1)}^{2}}$ .
Substitute 2 to obtain $\frac{8}{9}$ .
$\frac{8}{9}$
Differentiate $\frac{e^{x}}{x + 1}$ .
2 marks
The numerator is $e^{x} (x + 1) - e^{x} = x e^{x}$ .
Retain the excluded input.
$\frac{x e^{x}}{{(x + 1)}^{2}}$ , $x \neq - 1$
Find the tangent to $y = \frac{x + 1}{x - 1}$ at $x = 0$ .
3 marks
The point is $(0, - 1)$ .
The derivative is $- \frac{2}{{(x - 1)}^{2}}$ , so its value is $- 2$ .
Use point-gradient form.
$y = - 2 x - 1$
Why is the derivative formula for $\frac{x^{2} - 1}{x - 1}$ not valid at $x = 1$ , even though cancellation gives $x + 1$ elsewhere?
3 marks
Cancellation is valid only for $x \neq 1$ .
A derivative at an input requires the original function to be defined there.
The original function is undefined at 1.

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Curriculum reference & lesson scope

SACE Stage 2 Mathematical Methods

Further differentiation and applications from the current SACE Methods outline. Logarithmic derivatives have their own chapter; integration follows later. Trigonometric calculus uses radians.

Official curriculum source ↗

In this lesson

Understand the idea.

A ratio has two changing parts

Simplification and checking

See the method in action.

A rational derivative with its domain

Record the domain and factor derivatives.

Apply the quotient rule.

Simplify the whole numerator.

State the derivative with its restriction.

A tangent to a quotient

Find the contact point.

Differentiate the quotient.

Find the gradient at zero.

Write the tangent.

Try it yourself.

Check your understanding

Keep practising