The Product Rule
Why $(fg)' \neq f'g'$
When Leibniz first studied derivatives, he guessed that the derivative of a product would be the product of the derivatives: $(fg)' = f'g'$. He quickly discovered this was wrong.
Here’s a simple counterexample: Let $f(x) = x$ and $g(x) = x$. Then:
- $f'(x) = 1$ and $g'(x) = 1$, so $f'(x) \cdot g'(x) = 1$
- But $f(x)g(x) = x^2$, and $(x^2)' = 2x \neq 1$
The actual formula is more interesting, and once you understand it geometrically, it makes perfect sense.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Concept | Differentiation Formulas |
| Chapter | 2.3 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
Key Concepts
The Product Rule Formula
If $f$ and $g$ are both differentiable, then:
$$\boxed{\frac{d}{dx}[f(x)g(x)] = f(x) \cdot g'(x) + g(x) \cdot f'(x)}$$
In prime notation: $$(fg)' = fg' + gf'$$
In words: “First times the derivative of second, plus second times the derivative of first.”
Geometric Intuition: The Growing Rectangle
Imagine a rectangle with sides $f(x)$ and $g(x)$. Its area is $A = f(x) \cdot g(x)$.
g(x) Δg
┌──────────────┬─────┐
│ │ │
f(x)│ Original │ New │ Δf
│ Area │strip│
│ │ │
├──────────────┼─────┤
│ New strip │tiny │
└──────────────┴─────┘
f·Δg Δf·ΔgWhen both sides increase slightly:
- Original area: $f \cdot g$
- Horizontal strip (new area): $f \cdot \Delta g$ (first side times change in second)
- Vertical strip (new area): $g \cdot \Delta f$ (second side times change in first)
- Corner (tiny): $\Delta f \cdot \Delta g$ (negligible when both changes are small)
The rate of change of area is approximately: $$\frac{\Delta A}{\Delta x} \approx f \cdot \frac{\Delta g}{\Delta x} + g \cdot \frac{\Delta f}{\Delta x}$$
Taking the limit as $\Delta x \to 0$ gives us the Product Rule!
The Proof
Let $F(x) = f(x)g(x)$. Then:
$$F'(x) = \lim_{h \to 0}\frac{f(x+h)g(x+h) - f(x)g(x)}{h}$$
The trick: Add and subtract $f(x+h)g(x)$ in the numerator:
$$= \lim_{h \to 0}\frac{f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)}{h}$$
$$= \lim_{h \to 0}\left[f(x+h) \cdot \frac{g(x+h) - g(x)}{h} + g(x) \cdot \frac{f(x+h) - f(x)}{h}\right]$$
$$= f(x) \cdot g'(x) + g(x) \cdot f'(x)$$
(Note: $\lim_{h \to 0} f(x+h) = f(x)$ because $f$ is continuous.)
Memory Aids
Mnemonic 1: “First d-second plus second d-first”
- $(fg)' = f(g') + g(f')$
Mnemonic 2: Think of $f$ and $g$ as “partners”
- Each partner takes a turn being differentiated while the other stays the same
- Add the results
Common Mistake to Avoid
❌ Wrong: $(fg)' = f' \cdot g'$ (product of derivatives)
✓ Right: $(fg)' = fg' + gf'$ (sum of two terms)
Remember: If this were true, then $(x \cdot x)' = 1 \cdot 1 = 1$, but we know $(x^2)' = 2x$.
the derivative of a product is the product of the derivatives.
This is the multiplicative-not-additive error. Differentiation distributes over addition: $(f + g)' = f' + g'$ is true. It does NOT distribute over multiplication: $(f \cdot g)' = f' \cdot g'$ is false. The correct formula is additive -- a sum of two terms: $(fg)' = f \cdot g' + g \cdot f'$. The reason there are two terms is that both factors are changing simultaneously; when one changes, the other is still present. A concrete check: $f(x) = x^3$ and $g(x) = x^5$, so $(fg)(x) = x^8$ and $(fg)'(x) = 8x^7$. The wrong rule gives $f'g' = 3x^2 \cdot 5x^4 = 15x^6 \neq 8x^7$. The correct rule gives $f \cdot g' + g \cdot f' = x^3 \cdot 5x^4 + x^5 \cdot 3x^2 = 5x^7 + 3x^7 = 8x^7$.
the slope of $fg$ at a point equals the product of the slopes.
This is the height-vs-slope error applied to products. When you evaluate $(fg)'$ at a specific point, you must use the Product Rule formula, not multiply the two individual derivative values. At $x=2$: with $f(x) = x$ and $g(x) = x^4$, you have $f'(2) = 1$ and $g'(2) = 32$. The slope of the product $x^5$ at $x=2$ is $(x^5)'|_{x=2} = 5x^4|_{x=2} = 80$. The product of slopes $1 \cdot 32 = 32$ is wrong; the Product Rule gives $f(2)g'(2) + g(2)f'(2) = 2 \cdot 32 + 16 \cdot 1 = 64 + 16 = 80$, which is correct.
Practice Problems
Find $\frac{d}{dx}[(x^2)(x^3)]$ using the Product Rule.
Then verify by first multiplying and differentiating.
Differentiate $h(x) = (3x + 2)(x^2 - 1)$.
Find $f'(t)$ if $f(t) = \sqrt{t}(t^2 + 3t - 1)$.
Suppose $h(x) = xg(x)$, where $g(4) = 6$ and $g'(4) = -2$. Find $h'(4)$.
- Derive a formula for $(fgh)'$ where $f$, $g$, and $h$ are all differentiable functions.
- Use your formula to find the derivative of $y = x(x+1)(x+2)$.
CCI-Style Conceptual Questions
A company’s revenue is $R = p \cdot q$ where $p$ is price per unit and $q$ is quantity sold. At the current price, $p = 20$, $q = 1000$, $\frac{dp}{dt} = 0.50$ (price increasing), and $\frac{dq}{dt} = -30$ (quantity decreasing).
Is revenue increasing or decreasing? By how much per time unit?
For which functions $f$ is it true that $(xf(x))' = f(x) + xf'(x)$?
Mastery Checklist
Mental Model
Think of products as “partnerships”:
When two quantities are multiplied together and both change, the total rate of change has two contributions:
- How fast the first is changing (while second stays fixed)
- How fast the second is changing (while first stays fixed)
It’s like two people painting a wall together: the total progress depends on each person’s contribution while the other holds steady.
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|---|---|---|
| Sum/Difference Rules | Skills Index | Quotient Rule |
Last updated: 2026-01-22