Product Rule (HSC SSCE Mathematics Advanced): Revision Notes

Product Rule

The product rule is a differentiation technique used when you need to find the derivative of a function that is expressed as the product of two simpler functions multiplied together. This rule is essential when working with expressions that cannot be easily simplified by expanding.

Understanding the product rule

When you have a function that is the product of two separate functions, you cannot simply differentiate each part separately. Instead, you must use the product rule to find the derivative correctly.

For example, to differentiate $y = (x^3 + 4x + 2)(2x + 1)$, you recognise that this expression consists of two functions being multiplied together. The first factor, $(x^3 + 4x + 2)$, can be called $u(x)$ (or simply $u$), and the second factor, $(2x + 1)$, can be called $v(x)$ (or simply $v$).

The product rule formula

If you have a function $y = uv$ where both $u$ and $v$ are functions of $x$, then the derivative is:

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

Alternatively, using function notation, if $h(x) = u(x)v(x)$, then:

$h'(x) = u(x)v'(x) + v(x)u'(x)$

Components explained

Each part of the product rule formula has a specific meaning:

• $\frac{dy}{dx}$ is the derivative of the product $uv$
• $u$ is the first function
• $v$ is the second function
• $\frac{dv}{dx}$ is the derivative of the second function
• $\frac{du}{dx}$ is the derivative of the first function

Memory aid

How to apply the product rule

Follow these steps when using the product rule:

1. Identify the two functions: Determine which part of the expression is $u$ and which is $v$
2. Find the derivatives: Calculate $\frac{du}{dx}$ and $\frac{dv}{dx}$ separately
3. Apply the formula: Substitute into $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$
4. Simplify: Expand brackets and collect like terms
5. Factorise if possible: Often the result can be factorised to give a neater answer

Worked example 1: Simple product

Alternative approach:

For simpler products like this, it may be more efficient to expand the expression first:

$y = (3x - 1)(x + 5)$

$= 3x^2 + 15x - x - 5 = 3x^2 + 14x - 5$

Then differentiate term by term:

$\frac{dy}{dx} = \frac{d}{dx}(3x^2) + \frac{d}{dx}(14x) - \frac{d}{dx}(5)$

$= 6x + 14 - 0$

$= 6x + 14$

Both methods give the same result. For simpler products, expanding first can be more efficient. However, the product rule becomes essential when working with more complex functions.

Worked example 2: Product requiring chain rule

Choosing the right approach

When to expand first:

• If the product is simple and expanding creates a straightforward polynomial
• When expansion results in terms that are easy to differentiate

When to use the product rule:

• When expansion would be complicated or time-consuming
• When one or both functions involve composite functions requiring the chain rule
• When the product involves functions that cannot be easily multiplied out

Exam tips

• Always identify $u$ and $v$ clearly before starting
• Write out $\frac{du}{dx}$ and $\frac{dv}{dx}$ separately before applying the formula
• Remember to use other rules (like the chain rule or power rule) when finding the individual derivatives
• Check if expanding first might be simpler for basic products
• Factorise your final answer where possible

Remember!