Review of derivatives and polynomial rules | Year 12

01 / LEARN

Understand the method.

THE CORE RULE

\frac{d}{d x} (a x^{n}) = a n x^{n - 1}

Multiply by the exponent, then reduce it by one. Differentiate each term separately; a constant has derivative zero.

A height and a gradient are different values

The original function $f (a)$ gives the height of the curve. The derivative $f^{'} (a)$ gives the tangent gradient at the same input.

Find the derivative function first. Substitute the required input after differentiating.

\begin{aligned} Point & : (a, f (a)) \\ Gradient & : f^{'} (a) \end{aligned}

Rewrite roots and reciprocals as powers

The power rule works for real powers wherever the derivative exists. Rewrite a root or reciprocal before applying it.

Keep the domain: the real function $\sqrt{x}$ exists at $x = 0$ , but its derivative does not have a finite value there.

\begin{aligned} \sqrt{x} & = x^{\frac{1}{2}} \\ \frac{d}{d x} \sqrt{x} & = \frac{1}{2 \sqrt{x}} (x > 0) \end{aligned}

A tangent needs both the point and its gradient

Evaluate the original function for the point and the derivative for the gradient. Put both into the point–gradient equation.

If the horizontal input is time, the derivative measures a rate. For example, litres changing with seconds gives litres per second.

y - f (a) = f^{'} (a) (x - a)

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A gradient from a polynomial

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

Worked solution

1
Differentiate each term.
$f^{'} (x) = 6 x^{2} - 10 x + 4$
Keep each coefficient. The constant term disappears.
2
Substitute the required input.
$\begin{aligned} f^{'} (2) & = 6 {(2)}^{2} - 10 (2) + 4 \\ = 24 - 20 + 4 \\ = 8 \end{aligned}$
The result is the tangent gradient at $x = 2$ , not the height of the curve.

Answer

$f^{'} (2) = 8$ .

EXAMPLE 02

A tangent needs a point and a gradient

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

Worked solution

1
Find the point on the curve.
$f (2) = 2^{2} + 3 = 7$
The tangent passes through $(2, 7)$ .
2
Find the tangent gradient.
$\begin{aligned} f^{'} (x) & = 2 x \\ f^{'} (2) & = 4 \end{aligned}$
Use the derivative for the gradient and the original function for the point.
3
Write and simplify the tangent equation.
$\begin{aligned} y - 7 & = 4 (x - 2) \\ y & = 4 x - 1 \end{aligned}$
Substitution of $(2, 7)$ checks that the line passes through the required point.

Answer

$y = 4 x - 1$ .

03 / YOUR TURN

Try it yourself.

Calculate, explain, then apply.

CHECK 01

Differentiate a polynomial

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

Show answer & reasoning

$28 x^{3} - 3$ .

Multiply $7$ by the exponent $4$ , lower the power to $3$ , and differentiate the linear and constant terms.

Keep practising

6 worksheet questions · 15 marks

Open worksheet 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 $x^{- 2} + 4 x^{\frac{1}{2}}$ 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 $x^{- 2} + 4 x^{\frac{1}{2}}$ and state a real domain for the derivative.
3 marks
Differentiate the two powers.
The square-root derivative and reciprocal require $x > 0$ .
$- 2 x^{- 3} + 2 x^{- \frac{1}{2}}$ , 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 ↗

In this lesson

Understand the method.

A height and a gradient are different values

Rewrite roots and reciprocals as powers

A tangent needs both the point and its gradient

See the method in action.

A gradient from a polynomial

Differentiate each term.

Substitute the required input.

A tangent needs a point and a gradient

Find the point on the curve.

Find the tangent gradient.

Write and simplify the tangent equation.

Try it yourself.

Differentiate a polynomial

Keep practising