Product and quotient rules
A product or quotient can sometimes be simplified first. If you use these rules, differentiate u and v separately, then combine the contributions in the correct order
\[\frac{d}{dx}(uv)=u'v+uv',\quad \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}\]
- Name the structure Use the product rule for two factors multiplied together and the quotient rule for a numerator divided by a denominator \(\displaystyle f(x)=u(x)v(x)\quad\text{or}\quad f(x)=\frac{u(x)}{v(x)}\)
- Differentiate each factor Find the derivative of each factor separately before combining the results \(\displaystyle u=2x+1,\quad v=x^2+3,\quad u'=2,\quad v'=2x\)
- Use the product rule Keep one factor unchanged in each term and match each derivative with the other factor \(\displaystyle \frac{d}{dx}(uv)=u'v+uv'\)
- Use the quotient rule For a quotient, put v u' minus u v' over the square of the denominator and keep v non-zero \(\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2},\quad v\ne0\)
Worked example
Differentiate the quotient and find the value of its derivative at x = 0.
\[g(x)=\frac{x+1}{x^2+1}\]
- \(\displaystyle u(x)=x+1,\quad v(x)=x^2+1,\quad u'(x)=1,\quad v'(x)=2x\)Name the numerator and denominator, then differentiate each one
- \(\displaystyle g'(x)=\frac{\left(x^2+1\right)(1)-(x+1)(2x)}{\left(x^2+1\right)^2}\)Apply the quotient rule as v u' minus u v' over v squared
- \(\displaystyle g'(0)=\frac{(1)(1)-(1)(0)}{1^2}=\frac{1}{1}\)Substitute x = 0 into the derivative, not the original quotient
- \(\displaystyle g'(0)=1\)The exact derivative value is 1.
Common mistakes
Leaving out the factors
✕\(\displaystyle \frac{d}{dx}(uv)=u'+v'\)✓\(\displaystyle \frac{d}{dx}(uv)=u'v+uv'\)
The product rule keeps the undifferentiated factor in each term, so the derivatives are not added on their ownNot squaring the denominator
✕\(\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v}\)✓\(\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}\)
The quotient rule divides by the square of the original denominatorReversing the quotient numerator
✕\(\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{uv'-vu'}{v^2}\)✓\(\displaystyle \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}\)
The quotient-rule numerator is denominator times numerator derivative minus numerator times denominator derivativeDifficulty levels
introDifferentiate a product of a polynomial factor and a simple exponential or circular factor
standardDifferentiate a quotient of low-degree polynomial or power factors and keep the denominator condition
challengeRecognise an exact simplification or use a product or quotient derivative to form a tangent or normal gradient
examChoose the method and complete an exact derivative coefficient or value check for a combined function
Questions about this topic
How can the product rule and quotient rule be distinguished?+
Use the product rule when two factors are multiplied. Use the quotient rule when one factor is divided by another
Why is the denominator squared in the quotient rule?+
Write the quotient as the numerator multiplied by the reciprocal of the denominator. Apply the product and chain rules, then combine the terms over the denominator squared, provided the denominator is non-zero
When should a product or quotient be simplified first?+
Simplify first when exact algebra changes the expression into a simpler power or sum, while retaining any original domain restriction
Is my complete written rule automatically marked?+
No. Written rules and explanations are ungraded
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 6: Applications of differentiation
In this chapter
Differentiate polynomialsChallenge enrichment: rational powers and derivative domainsDifferentiate natural exponential functions and solve simple derivative conditionsDifferentiate sine, cosine and tangent functions and solve simple derivative conditionsDifferentiate a composite function by the chain ruleDifferentiate a product or quotient of two functionsEvaluate and determine function rules6B Differentiation rules: Find the maximal domain of an algebraic functionExpand and factor polynomial expressionsSimplify and combine algebraic fractions6B Differentiation rules: Find the equation of a line from its graph or from its intercept and slopeAnalyse polynomial curves, stationary points and extremaUse a derivative to analyse and recognise the graph of a function6C Tangents, normals and curve sketching: Evaluate a polynomial or recover a coefficient parameter6C Tangents, normals and curve sketching: Move between roots, factors, polynomial rules and equationsRecognise and classify a polynomialConnect a factorised higher-degree rule and its graphAnalyse a cubic graph or determine its ruleWork with sets and interval notation6C Tangents, normals and curve sketching: Find the maximal domain of an algebraic function6C Tangents, normals and curve sketching: Find the equation of a line from its graph or from its intercept and slopeForm an equation of a parallel or perpendicular lineSolve a maximum or minimum problem by differentiationSet up and solve a practical linear equationRearrange literal equations and recognise equivalent line forms6D Optimisation: Find the maximal domain of an algebraic function6D Optimisation: Evaluate a polynomial or recover a coefficient parameter6D Optimisation: Move between roots, factors, polynomial rules and equationsCalculate and interpret average and instantaneous rates from a modelVolume of cones, pyramids and spheresVolume of prisms and cylindersPerimeter and area of standard shapesSimilar figures and scale factorsFind a side of a right-angled triangleAnalyse straight-line motion through displacement, velocity and acceleration6F Kinematics: Evaluate a polynomial or recover a coefficient parameter