Derivative notation and the derivative function | Year 11

01 / LEARN

Understand the idea.

A function of gradients

The original function gives heights; its derivative gives tangent gradients. For each input where a derivative exists, $f^{'} (x)$ records the local rate there. Substituting $x = a$ into the derivative produces the number $f^{'} (a)$ , rather than the point height $f (a)$ .

When $y = f (x)$ , the symbols $y^{'}$ , $f^{'} (x)$ and $\frac{d y}{d x}$ describe the same derivative. Leibniz notation names both changing quantities, so $\frac{d s}{d t}$ makes clear that displacement changes with time. The prime is a derivative symbol, not an exponent.

\begin{array}{c} y = f (x) \\ ⟹ y^{'} = f^{'} (x) = \frac{d y}{d x} \end{array}

Meaning and dimensions

Derivative units come from output units divided by input units. For a temperature model in degrees Celsius and minutes, a derivative of $8 \circ C / \min$ means warming at that local rate. A negative derivative would mean cooling.

An instantaneous rate does not determine the current amount. Adding a constant to a function raises or lowers every height but leaves all tangent gradients unchanged. To recover a particular height from rates, additional information about the original function is needed.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

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

Worked solution

1
Differentiate the two terms.
$f^{'} (x) = 2 x + 3$
The gradient contributions from the quadratic and linear terms add.
2
Evaluate the derivative at input 2.
$f^{'} (2) = 2 (2) + 3 = 7$
Substitution into the derivative gives the tangent gradient at that input.
3
Evaluate the original function separately.
$f (2) = 2^{2} + 3 (2) = 10$
Substitution into the original function gives height 10, so the tangent point is $(2, 10)$ with gradient 7.

Answer

Derivative 7 at x=2; height 10.

EXAMPLE 02

Worked example 2

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

Worked solution

1
Differentiate temperature with respect to time.
$\frac{d T}{d t} = 6 t - 4$
The constant starting term contributes zero, while the two time-dependent terms give the rate function.
2
Evaluate the rate at 2 minutes.
${\frac{d T}{d t} |}_{t = 2} = 6 (2) - 4 = 8 \circ C / \min$
The derivative inherits temperature units divided by time units.
3
Interpret the positive rate.
${\frac{d T}{d t} |}_{t = 2} > 0$
At 2 minutes the model is warming at 8 degrees Celsius per minute; this is a local rate, not the temperature or a whole-minute change.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

11; the height cannot be determined.

$p^{'} (3) = 4 (3) - 1 = 11$ . This gives the local gradient only. Vertical shifts preserve the derivative, so $p (3)$ cannot be determined without an additional height condition.

Keep practising

6 worksheet questions · 18 marks

Open worksheet 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 $\frac{d V}{d t}$ .
3 marks
$A = π r^{2}$ and $\frac{d A}{d r} = 2 π r$ , 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 $8 x$ ; the constant contributes 0.
$\frac{d y}{d x} = 8 x$
$V$ is in litres and $t$ in minutes. State the units of $\frac{d V}{d t}$ .
3 marks
Divide the dependent-variable units by the independent-variable units.
Litres per minute
$A = π r^{2}$ and $\frac{d A}{d r} = 2 π r$ , 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 ↗

In this lesson

Understand the idea.

A function of gradients

Meaning and dimensions

See the method in action.

Worked example 1

Differentiate the two terms.

Evaluate the derivative at input 2.

Evaluate the original function separately.

Worked example 2

Differentiate temperature with respect to time.

Evaluate the rate at 2 minutes.

Interpret the positive rate.

Try it yourself.

Check your understanding

Keep practising