Product rule | Year 12 | Ascent Library

01 / LEARN

The idea, step by step

Why both factors matter

If $f (x) = u (x) v (x)$ , both factors can change with $x$ . Adding their two contributions gives $f^{'} = u^{'} v + u v^{'}$ . The product of the derivatives alone is not the derivative of the product.

Write the two factors on separate lines before differentiating. Keep brackets around each factor so signs and powers survive substitution.

Choose the efficient form

A polynomial product can be expanded first, providing an independent check. Keep an unexpanded form when it exposes roots or combines an exponential with a polynomial.

For a tangent, calculate the original function value as well as the derivative. The final answer may be algebraically equivalent in factored or expanded form.

For the exponential/trigonometric examples in this lesson, use the supplied facts ${(e^{x})}^{'} = e^{x}$ , ${(sin x)}^{'} = cos x$ and ${(cos x)}^{'} = - sin x$ for radian inputs. Their meaning is developed in the later exponential and trigonometric lessons.

A common trap

Differentiating $u v$ as $u^{'} v^{'}$ .

Instead: The product rule is a sum: change one factor at a time while retaining the other.

02 / WORKED EXAMPLES

See the thinking

Example 1

Two changing polynomial factors

Differentiate $y = (x^{2} + 1) (3 x - 2)$ .

Use $u^{'} = 2 x$ and $v^{'} = 3$ .
$y^{'} = 2 x (3 x - 2) + 3 (x^{2} + 1)$ .

9 x^{2} - 4 x + 3

Example 2

An exponential product at the origin

Find the tangent to $y = x e^{x}$ at $x = 0$ .

$y^{'} = e^{x} + x e^{x} = (1 + x) e^{x}$ .
The point is $(0, 0)$ and its gradient is 1.

y = x

Check your understanding

Differentiate $x^{2} sin x$ .

Show answer

$2 x sin x + x^{2} cos x$

Differentiate each factor in turn and add the two contributions.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Differentiate $(x + 2) (x^{2} - 3)$ .
2 marks
Differentiate $x^{3} e^{x}$ .
2 marks
Find $f^{'} (1)$ for $f (x) = (x^{2} + 2) (x^{2} - 1)$ .
2 marks
Differentiate $x cos x$ .
2 marks
Find the tangent to $y = (x^{2} + 1) (x + 1)$ at $x = 0$ .
3 marks
A rectangle has width $x$ cm and length $(10 - x)$ cm. Find the derivative of its area and explain why a product rule is relevant.
3 marks

Read the worked solutions

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

Differentiate $(x + 2) (x^{2} - 3)$ .
2 marks
Use $(x^{2} - 3) + (x + 2) 2 x$ .
Expand and collect terms.
$3 x^{2} + 4 x - 3$
Differentiate $x^{3} e^{x}$ .
2 marks
The two contributions are $3 x^{2} e^{x}$ and $x^{3} e^{x}$ .
Factor out $e^{x}$ .
$e^{x} (x^{3} + 3 x^{2})$
Find $f^{'} (1)$ for $f (x) = (x^{2} + 2) (x^{2} - 1)$ .
2 marks
The derivative is $2 x (x^{2} - 1) + 2 x (x^{2} + 2)$ .
At $x = 1$ , the first term is zero and the second is 6.
$6$
Differentiate $x cos x$ .
2 marks
Use $1 cos x + x (- sin x)$ .
Keep the negative sign from the cosine derivative.
$cos x - x sin x$
Find the tangent to $y = (x^{2} + 1) (x + 1)$ at $x = 0$ .
3 marks
The point is $(0, 1)$ .
The derivative is $2 x (x + 1) + (x^{2} + 1)$ , giving gradient 1.
Use the point-gradient equation.
$y = x + 1$
A rectangle has width $x$ cm and length $(10 - x)$ cm. Find the derivative of its area and explain why a product rule is relevant.
3 marks
The area is $x (10 - x)$ and both lengths change.
Differentiate to get $(10 - x) - x$ .
$A^{'} (x) = 10 - 2 x$

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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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