Combining Differentiation Rules (Product, Quotient, Chain)
Combining differentiation rules applies the product, quotient and chain rules together to differentiate products, quotients and compositions of functions.
Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Differential calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.
Theory
Combine the product, quotient and chain rules when functions are multiplied, divided or nested. This Year 12 Mathematics Advanced topic (MAV-12-04) applies all three rules.
Product: \(\dfrac{d}{dx}[uv]=u'v+uv'\). Quotient: \(\dfrac{d}{dx}\!\left[\dfrac{u}{v}\right]=\dfrac{u'v-uv'}{v^{2}}\). Chain: \(\dfrac{d}{dx}f(g(x))=f'(g(x))\,g'(x)\).
The first step is always to spot which structure you have.
Method
- Classify: product, quotient, or composite?
- Apply the matching rule, differentiating each part.
- Simplify the result.
| \(\dfrac{dy}{dx}\) | \(=\) | \((1)e^{x}+x e^{x}\) |
| \(=\) | \((1+x)e^{x}\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{(1)(x+1)-x(1)}{(x+1)^{2}}\) |
| \(=\) | \(\dfrac{1}{(x+1)^{2}}\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(5(2x+1)^{4}\cdot2\) |
| \(=\) | \(10(2x+1)^{4}\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(2x\sin x+x^{2}\cos x\) |
Common pitfalls
Frequently asked questions
What is the product rule?
The derivative of u times v is u prime v plus u v prime.
What is the quotient rule?
The derivative of u over v is (u prime v minus u v prime) all over v squared.
How do you know which differentiation rule to use?
Look at the structure: use the product rule for a product, the quotient rule for a fraction, and the chain rule for a function inside another function.
How do you differentiate x times e to the x?
By the product rule it is e to the x plus x e to the x, which factorises to (1 plus x) e to the x.