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The Product Rule

Reference: Stewart §2.3 • Chapter: 2 • Section: 3

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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·Δg

When 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 $(x^2)' = 2x$.

Practice Problems

Level 1 Basic Product

Find $\frac{d}{dx}[(x^2)(x^3)]$ using the Product Rule.

Then verify by first multiplying and differentiating.

Thought Process

Let $f(x) = x^2$ and $g(x) = x^3$. Apply the formula: $f \cdot g' + g \cdot f'$. To verify, note that $x^2 \cdot x^3 = x^5$, and differentiate directly.

Show Answer

Using the Product Rule:

Let $f(x) = x^2$ and $g(x) = x^3$.

Then $f'(x) = 2x$ and $g'(x) = 3x^2$.

$$(fg)' = f \cdot g' + g \cdot f' = x^2 \cdot 3x^2 + x^3 \cdot 2x = 3x^4 + 2x^5 = 5x^4$$

Verification:

$x^2 \cdot x^3 = x^5$, and $\frac{d}{dx}(x^5) = 5x^4$ ✓

Level 2 Polynomial Product

Differentiate $h(x) = (3x + 2)(x^2 - 1)$.

Thought Process

Let $f = 3x + 2$ and $g = x^2 - 1$. Find each derivative separately, then apply $fg' + gf'$.

Show Answer

Let $f(x) = 3x + 2$ and $g(x) = x^2 - 1$.

Then $f'(x) = 3$ and $g'(x) = 2x$.

$$h'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x)$$ $$= (3x + 2)(2x) + (x^2 - 1)(3)$$ $$= 6x^2 + 4x + 3x^2 - 3$$ $$= 9x^2 + 4x - 3$$

Level 3 Product with Radicals

Find $f'(t)$ if $f(t) = \sqrt{t}(t^2 + 3t - 1)$.

Thought Process

First rewrite $\sqrt{t} = t^{1/2}$. Let this be $f$ and let $g = t^2 + 3t - 1$. Apply the Product Rule, remembering that $\frac{d}{dt}(t^{1/2}) = \frac{1}{2}t^{-1/2}$.

Show Answer

Let $u(t) = t^{1/2}$ and $v(t) = t^2 + 3t - 1$.

Then $u'(t) = \frac{1}{2}t^{-1/2}$ and $v'(t) = 2t + 3$.

$$f'(t) = u \cdot v' + v \cdot u'$$ $$= t^{1/2}(2t + 3) + (t^2 + 3t - 1) \cdot \frac{1}{2}t^{-1/2}$$ $$= 2t^{3/2} + 3t^{1/2} + \frac{t^2 + 3t - 1}{2t^{1/2}}$$

To combine, multiply the first terms by $\frac{2t^{1/2}}{2t^{1/2}}$: $$= \frac{4t^2 + 6t}{2t^{1/2}} + \frac{t^2 + 3t - 1}{2t^{1/2}} = \frac{5t^2 + 9t - 1}{2\sqrt{t}}$$

Level 4 Using Function Values

Suppose $h(x) = xg(x)$, where $g(4) = 6$ and $g'(4) = -2$. Find $h'(4)$.

Thought Process

Apply the Product Rule to $h(x) = x \cdot g(x)$. Here $f(x) = x$, so $f'(x) = 1$. Then substitute $x = 4$ and use the given values.

Show Answer

Using the Product Rule with $f(x) = x$ and the given function $g(x)$:

$$h'(x) = x \cdot g'(x) + g(x) \cdot 1 = xg'(x) + g(x)$$

At $x = 4$: $$h'(4) = 4 \cdot g'(4) + g(4) = 4(-2) + 6 = -8 + 6 = -2$$

Level 5 Triple Product Rule

Derive a formula for $(fgh)'$ where $f$, $g$, and $h$ are all differentiable functions.
Use your formula to find the derivative of $y = x(x+1)(x+2)$.

Thought Process

For part (a): Apply the Product Rule twice by grouping. Let $F = fg$, so $(fgh) = F \cdot h$. Apply the Product Rule, then expand $F'$ using the Product Rule again.

For part (b): Apply your formula with $f = x$, $g = x+1$, $h = x+2$.

Show Answer

(a) Deriving the Triple Product Rule:

Let $F = fg$. Then $fgh = F \cdot h$.

By the Product Rule: $$(fgh)' = F'h + Fh' = (fg)'h + fgh'$$

Apply the Product Rule to $(fg)'$: $$= (f'g + fg')h + fgh' = f'gh + fg'h + fgh'$$

$$\boxed{(fgh)' = f'gh + fg'h + fgh'}$$

In words: Each function takes a turn being differentiated while the others stay the same.

(b) Applying to $y = x(x+1)(x+2)$:

Let $f = x$, $g = x+1$, $h = x+2$.

Then $f' = 1$, $g' = 1$, $h' = 1$.

$$y' = f'gh + fg'h + fgh'$$ $$= (1)(x+1)(x+2) + (x)(1)(x+2) + (x)(x+1)(1)$$ $$= (x+1)(x+2) + x(x+2) + x(x+1)$$ $$= (x^2 + 3x + 2) + (x^2 + 2x) + (x^2 + x)$$ $$= 3x^2 + 6x + 2$$

CCI-Style Conceptual Questions

Conceptual Economic Application

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?

Thought Process

Use the Product Rule: $\frac{dR}{dt} = p\frac{dq}{dt} + q\frac{dp}{dt}$. Substitute the values to find $\frac{dR}{dt}$. The sign tells us if revenue is increasing or decreasing.

Show Answer

By the Product Rule: $$\frac{dR}{dt} = p\frac{dq}{dt} + q\frac{dp}{dt}$$ $$= (20)(-30) + (1000)(0.50)$$ $$= -600 + 500 = -100$$

Revenue is decreasing by $100 per time unit.

Interpretation: The loss from selling fewer units (-$600) exceeds the gain from higher prices (+$500), so overall revenue falls.

Conceptual When Products Simplify

For which functions $f$ is it true that $(xf(x))' = f(x) + xf'(x)$?

Thought Process

Apply the Product Rule to $(xf(x))'$ in general and compare to the given expression.

Show Answer

Using the Product Rule on $xf(x)$ with first function $x$ and second function $f(x)$:

$$(xf(x))' = x \cdot f'(x) + f(x) \cdot 1 = xf'(x) + f(x) = f(x) + xf'(x)$$

This is always true for any differentiable function $f$! The given formula is simply the Product Rule applied to $xf(x)$.

Mastery Checklist

[ ] I can state and apply the Product Rule formula
[ ] I understand the geometric interpretation (growing rectangle)
[ ] I can differentiate products without multiplying out first
[ ] I can handle products involving radicals and negative exponents
[ ] I can use the Product Rule with function values (like $f(a)$ and $f'(a)$)
[ ] I can extend to products of three functions

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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Last updated: 2026-01-22