Product rule | Year 12 | Ascent Library

01 / LEARN

Understand the idea.

Why both factors matter

A product has two factors changing at once. For $y = u (x) v (x)$ , each contributes to the total rate: $y^{'} = u^{'} v + u v^{'}$ . In each term, differentiate one factor and keep the other unchanged.

Identify the two complete factors before calculating their derivatives. Retaining brackets until both product-rule terms are written helps prevent a lost constant, sign or power.

Choose the efficient form

Expanding a polynomial product before differentiating provides an independent check. With an exponential or trigonometric factor, the product form usually makes the two contributions easier to track. A tangent also needs the point from the original function.

Use the supplied facts ${(e^{x})}^{'} = e^{x}$ , ${(sin x)}^{'} = cos x$ and ${(cos x)}^{'} = - sin x$ . The trigonometric rules here use radians; these facts are developed in later lessons.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Two changing polynomial factors

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

Worked solution

1
Identify and differentiate the factors.
$\begin{aligned} u & = x^{2} + 1, & u^{'} & = 2 x \\ v & = 3 x - 2, & v^{'} & = 3 \end{aligned}$
The constant term in each factor differentiates to zero.
2
Add the two product contributions.
$y^{'} = 2 x (3 x - 2) + 3 (x^{2} + 1)$
Each changing factor contributes while the other remains unchanged.
3
Expand and collect like terms.
$\begin{aligned} y^{'} & = 6 x^{2} - 4 x + 3 x^{2} + 3 \\ = 9 x^{2} - 4 x + 3 \end{aligned}$
The two quadratic terms combine; the constant 3 remains.

Answer

$9 x^{2} - 4 x + 3$

EXAMPLE 02

An exponential product at the origin

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

Worked solution

1
Find the point on the original curve.
$y (0) = 0 e^{0} = 0$
The tangent passes through the curve at the stated input.
2
Differentiate both factors in turn.
$\begin{aligned} y^{'} & = 1 \cdot e^{x} + x e^{x} \\ = (1 + x) e^{x} \end{aligned}$
Both x and the exponential vary with x.
3
Evaluate the tangent gradient.
$y^{'} (0) = (1 + 0) e^{0} = 1$
The derivative at the contact point gives the line gradient.
4
Form the tangent equation.
$y - 0 = 1 (x - 0) ⟹ y = x$
Point-gradient form uses the point (0,0) and gradient 1.

Answer

$y = x$

03 / YOUR TURN

Try it yourself.

Check your understanding

Differentiate $x^{2} sin x$ .

Show answer & reasoning

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

Take $u = x^{2}$ and $v = sin x$ . The two contributions are $2 x sin x$ and $x^{2} cos x$ , giving $2 x sin x + x^{2} cos x$ for radian input.

Keep practising

6 worksheet questions · 14 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Why both factors matter

Choose the efficient form

See the method in action.

Two changing polynomial factors

Identify and differentiate the factors.

Add the two product contributions.

Expand and collect like terms.

An exponential product at the origin

Find the point on the original curve.

Differentiate both factors in turn.

Evaluate the tangent gradient.

Form the tangent equation.

Try it yourself.

Check your understanding

Keep practising