Derivative notation and the derivative function | Year 11

01 / LEARN

The idea, step by step

A function of gradients

The derivative function f′(x) assigns a tangent gradient to each input where that gradient exists. Once x=a is specified, f′(a) is a number. It is different from the height f(a).

If y depends on x, dy/dx is another notation for the derivative. For displacement depending on time, ds/dt makes the independent variable clear. Choose notation that matches the quantities in the model.

y = f (x) ⟹ y^{'} = f^{'} (x) = \frac{d y}{d x} .

Meaning and dimensions

A derivative has output units divided by input units. Temperature in degrees Celsius against minutes gives degrees Celsius per minute. Its sign tells whether the model is locally warming or cooling.

The prime does not mean multiplication or squaring. Knowing f′(2)=5 tells you a local rate, not the value of f(2). Different functions can have identical derivative functions but different heights.

A common trap

Using a derivative value as the height of the tangent point.

Instead: The point is (a,f(a)); its gradient is f′(a). Calculate them separately.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

For $f (x) = x^{2} + 3 x$ , find $f^{'} (x)$ , $f^{'} (2)$ and $f (2)$ .

$f^{'} (x) = 2 x + 3$ .
$f^{'} (2) = 7$ .
$f (2) = 4 + 6 = 10$ .

Derivative 7 at x=2; height 10.

Example 2

Worked example 2

$T (t) = 20 + 3 t^{2} - 4 t$ degrees Celsius, where t is in minutes. Interpret dT/dt at 2.

$d T / d t = 6 t - 4$ .
At t=2 the derivative is 8.
The units are degrees Celsius per minute.

At 2 minutes the model is warming at 8°C per minute.

Check your understanding

$p^{'} (x) = 4 x - 1$ . Find $p^{'} (3)$ . Can you also determine $p (3)$ ?

Show answer

11; the height cannot be determined.

$4 (3) - 1 = 11$ . An additional height condition is needed.

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

If $f^{'} (x) = 2 x + 5$ , calculate $f^{'} (- 2)$ and interpret it.
3 marks
For $y = 4 x^{2} - 1$ , write the derivative using Leibniz notation.
3 marks
V is in litres and t in minutes. State the units of $d V / d t$ .
3 marks
$A = π r^{2}$ and $d A / d r = 2 π r$ , with r in cm. Evaluate the derivative at 3 and give units.
3 marks
For $f (x) = x^{2} + 3 x$ , find $f (1)$ and $f^{'} (1)$ .
3 marks
A student writes $f^{'} (2) = f {(2)}^{2}$ . Explain the error.
3 marks

Read the worked solutions

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

If $f^{'} (x) = 2 x + 5$ , calculate $f^{'} (- 2)$ and interpret it.
3 marks
$2 (- 2) + 5 = 1$ .
It is a slope, not a height.
1; tangent gradient at x=−2.
For $y = 4 x^{2} - 1$ , write the derivative using Leibniz notation.
3 marks
The derivative of $4 x^{2}$ is 8x; the constant contributes 0.
$d y / d x = 8 x$
V is in litres and t in minutes. State the units of $d V / d t$ .
3 marks
Divide the dependent-variable units by the independent-variable units.
Litres per minute
$A = π r^{2}$ and $d A / d r = 2 π r$ , with r in cm. Evaluate the derivative at 3 and give units.
3 marks
Substitute r=3 to obtain $6 π$ .
Area-change units divided by radius-change units are cm²/cm.
$6 π$ cm² per cm
For $f (x) = x^{2} + 3 x$ , find $f (1)$ and $f^{'} (1)$ .
3 marks
$f (1) = 1 + 3 = 4$ .
$f^{'} (x) = 2 x + 3$ , so $f^{'} (1) = 5$ .
4 and 5
A student writes $f^{'} (2) = f {(2)}^{2}$ . Explain the error.
3 marks
One operation measures a rate and the other squares an output value; they need not agree.
The prime denotes differentiation, not squaring.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topic 6: rates of change, derivatives and applications. Introductory polynomial calculus with explicit modelling and interpretation.

Official curriculum source ↗

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