Differentiating powers and polynomials | Year 11

01 / LEARN

Understand the idea.

The power rule

For a polynomial term, multiply the coefficient by its positive integer exponent, then reduce the exponent by one. For example, $5 x^{3}$ contributes $15 x^{2}$ to the derivative. Carry the sign of each term through the calculation.

A linear term such as $7 x$ has constant gradient 7. A constant such as $- 9$ changes no heights relative to neighbouring inputs, so its derivative is zero. Differentiate each term separately and then collect the results.

\begin{array}{c} \frac{d}{d x} (a x^{n}) = a n x^{n - 1} (n \geq 1) \\ \frac{d}{d x} (c) = 0 \end{array}

Simplify before differentiating

If an expression is presented as a product of brackets, expand it first to obtain a polynomial. Collect like terms before differentiating. The derivative of a product is generally not the product of the two derivatives, so expansion avoids that trap here.

For a nonconstant polynomial, the derivative has degree one less than the original. This provides a useful first check, but it will not detect every coefficient error. Check the multiplier and reduced exponent separately, including any negative coefficients.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Differentiate $f (x) = 4 x^{3} - 5 x^{2} + 7 x - 9$ .

Worked solution

1
Differentiate the cubic and quadratic terms.
$\begin{aligned} \frac{d}{d x} (4 x^{3}) & = 4 (3) x^{2} = 12 x^{2} \\ \frac{d}{d x} (- 5 x^{2}) & = - 5 (2) x = - 10 x \end{aligned}$
Each exponent becomes a multiplier before decreasing by one, and the negative coefficient stays negative.
2
Differentiate the linear and constant terms.
$\begin{aligned} \frac{d}{d x} (7 x) & = 7 \\ \frac{d}{d x} (- 9) & = 0 \end{aligned}$
A line has constant gradient; a fixed vertical offset has zero rate of change.
3
Add the term derivatives.
$f^{'} (x) = 12 x^{2} - 10 x + 7$
Differentiation preserves sums and differences, giving a quadratic gradient function for this cubic.

Answer

$f^{'} (x) = 12 x^{2} - 10 x + 7$ .

EXAMPLE 02

Worked example 2

Differentiate $y = (x + 2) (x - 3)$ .

Worked solution

1
Expand the product.
$\begin{aligned} y & = (x + 2) (x - 3) \\ = x^{2} - 3 x + 2 x - 6 \\ = x^{2} - x - 6 \end{aligned}$
An equivalent polynomial lets us apply the term-by-term power rule.
2
Differentiate each polynomial term.
$\begin{aligned} \frac{d}{d x} (x^{2}) & = 2 x \\ \frac{d}{d x} (- x) & = - 1 \\ \frac{d}{d x} (- 6) & = 0 \end{aligned}$
The quadratic, linear and constant terms contribute their own gradients.
3
Combine the results.
$\frac{d y}{d x} = 2 x - 1$
The constant term disappears and the derivative is linear, one degree below the original quadratic.

Answer

$d y / d x = 2 x - 1$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Differentiate $7 x^{4} - 2 x$ .

Show answer & reasoning

$28 x^{3} - 2$

$\frac{d}{d x} (7 x^{4}) = 28 x^{3}$ and $\frac{d}{d x} (- 2 x) = - 2$ , so the derivative is $28 x^{3} - 2$ . The coefficient is multiplied by the old exponent in each nonconstant term.

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

Differentiate $3 x^{4} - 2 x^{2} + 5$ .
3 marks
Find $\frac{d}{d t}$ of $6 t^{3} + t - 8$ .
3 marks
Differentiate $y = 9 - 4 x$ .
3 marks
Differentiate $(x + 2) (x - 3)$ .
3 marks
Differentiate $f (x) = \frac{1}{2} x^{4} - 3 x^{2} + 6 x$ .
3 marks
A student differentiates $8 x^{3}$ as $8 x^{2}$ . Correct and explain.
3 marks

Read the worked solutions

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

Differentiate $3 x^{4} - 2 x^{2} + 5$ .
3 marks
Differentiate each power; the constant 5 contributes 0.
$12 x^{3} - 4 x$
Find $\frac{d}{d t}$ of $6 t^{3} + t - 8$ .
3 marks
$6 (3) t^{2} + 1 - 0 = 18 t^{2} + 1$ .
$18 t^{2} + 1$
Differentiate $y = 9 - 4 x$ .
3 marks
The derivative of 9 is 0; that of $- 4 x$ is $- 4$ .
$- 4$
Differentiate $(x + 2) (x - 3)$ .
3 marks
Expand to $x^{2} - x - 6$ , then differentiate.
$2 x - 1$
Differentiate $f (x) = \frac{1}{2} x^{4} - 3 x^{2} + 6 x$ .
3 marks
$\frac{1}{2} (4) x^{3} - 3 (2) x + 6 = 2 x^{3} - 6 x + 6$ .
$2 x^{3} - 6 x + 6$
A student differentiates $8 x^{3}$ as $8 x^{2}$ . Correct and explain.
3 marks
Multiply the coefficient by 3 while reducing the exponent to 2.
$24 x^{2}$

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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.

The power rule

Simplify before differentiating

See the method in action.

Worked example 1

Differentiate the cubic and quadratic terms.

Differentiate the linear and constant terms.

Add the term derivatives.

Worked example 2

Expand the product.

Differentiate each polynomial term.

Combine the results.

Try it yourself.

Check your understanding

Keep practising