How do we differentiate functions that are products or quotients of simpler functions?
The product and quotient rules differentiate functions formed by multiplying or dividing two differentiable functions.
How to use the product rule and quotient rule to differentiate functions built by multiplying or dividing two functions, with full worked examples.
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What this dot point is asking
You need to recognise when a function is a product of two functions or a quotient of two functions, choose the correct rule, and apply it carefully. These rules build on the basic derivatives of polynomials, exponentials, logarithms and trig functions, so accuracy in those is assumed.
The product rule
If where and are both differentiable, then
A useful shorthand is "first times derivative of second, plus second times derivative of first" - but written as the order does not matter because addition is commutative.
When to use it
Use the product rule whenever the function is a multiplication of two non-trivial expressions, for example , , or . If one factor is a constant you do not need the product rule - just keep the constant.
The quotient rule
If where and are differentiable and , then
The order in the numerator does matter here because of the subtraction: it is always (derivative of top)(bottom) minus (top)(derivative of bottom), all over (bottom) squared.
Choosing the right rule
- A product (two things multiplied): product rule.
- A quotient (one thing divided by another): quotient rule, or rewrite as a product with a negative power and use the product/chain rule.
- A sum or difference: differentiate term by term - no special rule needed.
Sometimes a quotient is easier as a product. For example can be differentiated with the product rule, which some students find less error-prone than the quotient rule.
Common errors
Bringing it together
The product and quotient rules are the workhorses of Stage 2 differentiation. They combine constantly with the chain rule (the next dot point) - for instance needs the product rule on the outside and the chain rule to differentiate . Master identifying and cleanly, write the rule before substituting, and factorise your answer where possible so it is ready for the applications that follow.
Exam-style practice questions
Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2023 SACE Stage 22 marksFind dy/dx for y = ln(5x^2) / x^2. There is no need to simplify your answer.Show worked answer →
This is a quotient, so use the quotient rule with u = ln(5x^2) and v = x^2.
First the derivatives of the parts. By the chain rule, u' = (d/dx) ln(5x^2) = (10x) / (5x^2) = 2/x. The denominator gives v' = 2x.
Apply the quotient rule (u'v - uv') / v^2:
dy/dx = ( (2/x)(x^2) - ln(5x^2)(2x) ) / (x^2)^2 = ( 2x - 2x ln(5x^2) ) / x^4.
The question says there is no need to simplify, so full marks go to a correct unsimplified quotient-rule expression. One mark is for differentiating ln(5x^2) to 2/x, and one mark for the correct quotient-rule structure.
2018 SACE Stage 23 marksDetermine dy/dx for y = (e^x + e^(-x)) / x. You do not need to simplify your answer.Show worked answer →
Use the quotient rule with u = e^x + e^(-x) and v = x.
Differentiate the parts: u' = e^x - e^(-x) (the chain rule sends e^(-x) to -e^(-x)), and v' = 1.
Quotient rule (u'v - uv') / v^2:
dy/dx = ( (e^x - e^(-x)) x - (e^x + e^(-x))(1) ) / x^2.
Marks: one for u' with the correct sign on e^(-x), one for v' = 1 with the quotient-rule set-up, and one for a correctly assembled final expression. No simplification is required.