Quotient rule | Year 12 | Ascent Library

01 / LEARN

The idea, step by step

A ratio has two changing parts

For $y = u / v$ , with $v \neq 0$ , the quotient rule gives $y^{'} = (u^{'} v - u v^{'}) / v^{2}$ . The subtraction order matters: derivative of the top times the bottom, minus the top times derivative of the bottom.

Write the domain before beginning. Squaring the denominator in the derivative does not make excluded inputs valid.

Simplification and checking

If division separates an expression into simpler powers, use that before differentiating. For example, $(x^{2} + 1) / x = x + x^{- 1}$ for $x \neq 0$ .

Equivalent formulas can look different. Expand or factor the derivative numerator to check equivalence, rather than judging only its appearance.

For the exponential/trigonometric examples in this lesson, use the supplied facts ${(e^{x})}^{'} = e^{x}$ , ${(sin x)}^{'} = cos x$ and ${(cos x)}^{'} = - sin x$ for radian inputs. Their meaning is developed in the later exponential and trigonometric lessons.

A common trap

Reversing the subtraction in the numerator.

Instead: Keep $u^{'} v - u v^{'}$ in that order and put the whole numerator over $v^{2}$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

A rational derivative with its domain

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

$u^{'} = 2 x$ , $v^{'} = 1$ .
$y^{'} = [2 x (x - 1) - (x^{2} + 1)] / {(x - 1)}^{2}$ .

\frac{x^{2} - 2 x - 1}{{(x - 1)}^{2}}

x \neq 1

Example 2

A tangent to a quotient

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

$y^{'} = [(x + 2) - x] / {(x + 2)}^{2} = 2 / {(x + 2)}^{2}$ .
The point is $(0, 0)$ and the gradient is $1 / 2$ .

y = x / 2

Check your understanding

Differentiate $1 / (x + 3)$ .

Show answer

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

The numerator is $0 (x + 3) - 1 (1)$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Differentiate $(3 x + 1) / (x + 2)$ .
2 marks
Differentiate $(x^{2} - 4) / x$ efficiently.
2 marks
Find $f^{'} (2)$ if $f (x) = x^{2} / (x + 1)$ .
2 marks
Differentiate $e^{x} / (x + 1)$ .
2 marks
Find the tangent to $y = (x + 1) / (x - 1)$ at $x = 0$ .
3 marks
Why is the derivative formula for $(x^{2} - 1) / (x - 1)$ 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 $(3 x + 1) / (x + 2)$ .
2 marks
The numerator is $3 (x + 2) - (3 x + 1) = 5$ .
Divide by ${(x + 2)}^{2}$ .
$5 / {(x + 2)}^{2}$ , $x \neq - 2$
Differentiate $(x^{2} - 4) / x$ efficiently.
2 marks
Rewrite as $x - 4 x^{- 1}$ .
Use the power rule.
$1 + 4 / x^{2}$ , $x \neq 0$
Find $f^{'} (2)$ if $f (x) = x^{2} / (x + 1)$ .
2 marks
The derivative is $[2 x (x + 1) - x^{2}] / {(x + 1)}^{2}$ .
Substitute 2 to obtain $8 / 9$ .
$8 / 9$
Differentiate $e^{x} / (x + 1)$ .
2 marks
The numerator is $e^{x} (x + 1) - e^{x} = x e^{x}$ .
Retain the excluded input.
$x e^{x} / {(x + 1)}^{2}$ , $x \neq - 1$
Find the tangent to $y = (x + 1) / (x - 1)$ at $x = 0$ .
3 marks
The point is $(0, - 1)$ .
The derivative is $- 2 / {(x - 1)}^{2}$ , so its value is $- 2$ .
Use point-gradient form.
$y = - 2 x - 1$
Why is the derivative formula for $(x^{2} - 1) / (x - 1)$ 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.

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