Short on time? This page is the recap. For the area picture, the proof and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: f′(x) and d/dx f(x) both mean the derivative of f.
Recap
- For a product of two differentiable functions: (f · g)′ = f′ · g + f · g′.
- In words: differentiate one factor at a time and add the two results.
- The derivative of a product is not the product of the derivatives.
- Why it works: a rectangle with sides f and g grows by two strips, g · Δf and f · Δg, plus a corner that becomes negligible.
- The proof subtracts and adds f(x + h)g(x), and uses the fact that differentiable functions are continuous.
- A constant factor needs no product rule: (c · g)′ = c · g′.
Key relationships
| Situation | Derivative |
|---|---|
| f(x) · g(x) | f′(x)g(x) + f(x)g′(x) |
| c · g(x), c constant | c · g′(x) |
| u · v · w | u′vw + uv′w + uvw′ |
| Table at x = a | f′(a)g(a) + f(a)g′(a) |
| x · ln x | ln x + 1 |
| xⁿ · eˣ | n·x^(n − 1)·eˣ + xⁿ·eˣ = eˣ·x^(n − 1)(n + x) |
Assumptions behind the rule
- Both factors must be differentiable at the point.
- Each factor’s input is plain x here. Compositions such as sin(x²) need the chain rule (Topic 3.1).
- Quotients need the quotient rule (Topic 2.9).
Mistakes to avoid
- Writing f′ · g′. Check: for x² · x³ = x⁵ the derivative is 5x⁴, not 2x · 3x² = 6x³.
- Dropping a term by treating one factor as a constant.
- Losing the minus sign from d/dx cos x = −sin x inside the rule.
- Using f(a)g(a) (the value) when the question asks for the slope.
- Not factoring a common factor such as eˣ before solving y′ = 0.
Quick self-check
- Differentiate x³eˣ. (3x²eˣ + x³eˣ = x²(x + 3)eˣ)
- Differentiate sin x cos x. (cos x · cos x + sin x · (−sin x) = cos²x − sin²x)
- f(1) = 4, f′(1) = 2, g(1) = −3, g′(1) = 1. Find (f · g)′(1). (2(−3) + 4(1) = −2)
- Do you need the product rule for 5 ln x? (No: it is a constant multiple, so the derivative is 5/x.)
Next: practice questions.