The Product Rule
“Product rule: differentiate one at a time, then add.”
The formula
(f·g)' = f'·g + f·g'How to read it: the derivative of a product is 'differentiate the first, keep the second' plus 'keep the first, differentiate the second'
- f, g
- — the two functions being multiplied
- f', g'
- — the derivative of each function
- f'g + fg'
- — 'differentiate the first × keep the second' + 'keep the first × differentiate the second'
The hook
The most common blunder is believing (fg)' = f'g' — the derivative of a product isn't that simple.
In plain words
To differentiate a product of two functions, differentiate one at a time and add the two results.
The intuition
Picture a rectangle with width f and height g, area fg. Nudge each side a little and the area grows two ways — by however much the width grew (f'g) and by however much the height grew (fg'). Their sum is the area's rate of growth.
How it's built
Two pieces — f'g (the contribution when only the first changes) and fg' (when only the second changes). The tiny sliver where both change at once (f'g'·h) vanishes as h→0, so it doesn't survive.
Example
With f=x², g=x³: f'g+fg' = 2x·x³ + x²·3x² = 2x⁴+3x⁴ = 5x⁴. And indeed x²·x³=x⁵ differentiates to 5x⁴ — a perfect match.
Common mistake
(fg)' ≠ f'g'. Differentiating each and multiplying is wrong. You must ADD the two terms.
Where it's used
Anywhere a product of two changing quantities shows up — speed×mass, length×width. It's also a building block for the derivatives that come later.
Where it came from
Leibniz worked out the derivative of a product in the 1670s, setting down the rule in the form we use today.
Prerequisites
Quick check
What does (f·g)' equal?
- f'g'
- f'g + fg'✓
- f' + g'
- fg
Practice
Differentiate h(x)=x·x² with the product rule and find h'(2).
Answer: 12
- f=x, f'=1 ; g=x², g'=2x
- h'=1·x² + x·2x = x²+2x² = 3x²
- Plug in x=2 → 3·4 = 12
Takeaway: Same as differentiating x·x²=x³ to 3x² — the check passes.
Differentiate h(x)=(x+1)(x+2) with the product rule and find h'(1).
Answer: 5
- f=x+1, f'=1 ; g=x+2, g'=1
- h'=(x+2) + (x+1) = 2x+3
- Plug in x=1 → 5
Takeaway: Expanded it's x²+3x+2, derivative 2x+3 — results agree.
Differentiate h(x)=x²·(x+1) with the product rule, then expand to double-check.
Answer: undefined
- f=x², f'=2x ; g=x+1, g'=1
- h'=2x(x+1) + x²·1 = 2x²+2x+x² = 3x²+2x
- Check: expand to x³+x², differentiate to 3x²+2x — matches
Takeaway: Product rule and expand-then-differentiate always agree.