Review of derivatives and polynomial rules | Year 12

01 / LEARN

The idea, step by step

What the derivative measures

For a curve $y = f (x)$ , the value $f^{'} (a)$ is the gradient of its tangent at $x = a$ . The derivative function gives this gradient at each allowed input, rather than giving the height of the curve.

If $x$ is time in seconds and $f (x)$ is distance in metres, the derivative has units of metres per second. Always name which quantity is changing.

Power and sum rules

For a constant real exponent $n$ , differentiate $x^{n}$ as $n x^{n - 1}$ wherever the original function and derivative are defined. Constants differentiate to zero. Keep coefficients and differentiate a sum term by term.

Rewrite roots and reciprocals as powers when helpful. For example, $\sqrt{x} = x^{1 / 2}$ has derivative $1 / (2 \sqrt{x})$ for $x > 0$ ; the endpoint $x = 0$ needs separate consideration.

A common trap

Using the derivative value as the y-coordinate.

Instead: Evaluate the original function for the point and the derivative for its gradient.

02 / WORKED EXAMPLES

See the thinking

Example 1

A gradient from a polynomial

$f (x) = 2 x^{3} - 5 x^{2} + 4 x - 7$ . Find $f^{'} (2)$ .

Differentiate each term: $f^{'} (x) = 6 x^{2} - 10 x + 4$ .
Substitute $x = 2$ : $24 - 20 + 4 = 8$ .

8

Example 2

A tangent needs a point and a gradient

Find the tangent to $y = x^{2} + 3$ at $x = 2$ .

The point is $(2, 7)$ and the derivative is $2 x$ .
The tangent gradient is $4$ , so $y - 7 = 4 (x - 2)$ .

y = 4 x - 1

Check your understanding

Differentiate $7 x^{4} - 3 x + 8$ .

Show answer

$28 x^{3} - 3$

The constant has derivative zero; the power rule multiplies by 4 and lowers the power by 1.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Differentiate $3 x^{5} - 2 x^{2} + 9$ .
2 marks
Find the gradient of $y = x^{3} - 4 x$ at $x = - 1$ .
2 marks
Differentiate $x^{- 2} + 4 x^{1 / 2}$ and state a real domain for the derivative.
3 marks
Find a tangent equation to $y = 2 x^{2} - x$ at $x = 1$ .
3 marks
A water-volume model is $V (t) = 50 + 6 t - t^{2}$ litres, for $0 \leq t \leq 3$ seconds. Find and interpret $V^{'} (2)$ .
3 marks
A student says the derivative of $5 x^{3}$ is $5 x^{2}$ . Explain the error and correct it.
2 marks

Read the worked solutions

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

Differentiate $3 x^{5} - 2 x^{2} + 9$ .
2 marks
Apply the power rule to both variable terms.
The derivative of 9 is zero.
$15 x^{4} - 4 x$
Find the gradient of $y = x^{3} - 4 x$ at $x = - 1$ .
2 marks
The derivative is $3 x^{2} - 4$ .
At $x = - 1$ , this is $3 - 4 = - 1$ .
$- 1$
Differentiate $x^{- 2} + 4 x^{1 / 2}$ and state a real domain for the derivative.
3 marks
Differentiate the two powers.
The square-root derivative and reciprocal require $x > 0$ .
$- 2 x^{- 3} + 2 x^{- 1 / 2}$ , for $x > 0$ .
Find a tangent equation to $y = 2 x^{2} - x$ at $x = 1$ .
3 marks
The point is $(1, 1)$ .
The derivative $4 x - 1$ gives gradient 3.
Use $y - 1 = 3 (x - 1)$ .
$y = 3 x - 2$
A water-volume model is $V (t) = 50 + 6 t - t^{2}$ litres, for $0 \leq t \leq 3$ seconds. Find and interpret $V^{'} (2)$ .
3 marks
Differentiate to get $V^{'} (t) = 6 - 2 t$ .
At 2 seconds, the rate is $2$ L/s.
$2$ litres per second: volume is increasing.
A student says the derivative of $5 x^{3}$ is $5 x^{2}$ . Explain the error and correct it.
2 marks
The exponent 3 must multiply the coefficient.
Lowering the exponent alone omits this factor.
$15 x^{2}$

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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 ↗

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