Product and Quotient Rules

When you learned multiplication as a child, you probably discovered a satisfying pattern: to multiply 12 by 3, you could break it into (10 + 2) times 3 and add the pieces. Math felt predictable. So when you first encounter derivatives, you might naturally assume that the derivative of a product is just the product of the derivatives. It would be elegant. It would be simple. And it would be completely wrong.

This lesson is about why that intuition fails and what actually happens when you differentiate products and quotients of functions. The good news is that the correct rules, while not as simple as you might hope, are remarkably elegant in their own way. Once you understand why they work, you will never forget them. And once you have practiced them, they will become as automatic as that multiplication trick you learned years ago.

Think of it this way: when two quantities that are both changing get multiplied together, the result changes in a more complex way than if just one were changing. A rectangle whose length and width are both growing expands faster than one where only the length grows. This everyday intuition is exactly what the product rule captures.

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Core Concepts

Why the Simple Approach Fails

Let us address the elephant in the room immediately. If $f(x)$ and $g(x)$ are both functions of $x$, you might hope that:

$$(f \cdot g)’ = f’ \cdot g’$$

This is false, and understanding why will help you remember the correct rule.

Consider a simple example: let $f(x) = x$ and $g(x) = x$. Then $f(x) \cdot g(x) = x^2$.

If the simple rule worked, we would have:

• $f’(x) = 1$ and $g’(x) = 1$
• So $(fg)’ = f’ \cdot g’ = 1 \cdot 1 = 1$

But we know that $\frac{d}{dx}[x^2] = 2x$, not 1!

The “naive” rule gives us a constant when we should get a function of $x$. Something is clearly missing. What is missing is the interaction between the two changing quantities.

The Product Rule: Statement and Intuition

The correct rule for differentiating a product is:

$$\boxed{(fg)’ = f’g + fg’}$$

Or, using Leibniz notation:

$$\frac{d}{dx}[f(x) \cdot g(x)] = f’(x) \cdot g(x) + f(x) \cdot g’(x)$$

In words: the derivative of a product equals the first function’s derivative times the second, plus the first times the second’s derivative.

Some people remember this as: “derivative of the first times the second, plus the first times the derivative of the second.”

The Rectangle Area Analogy

Here is the intuition that makes this rule unforgettable. Imagine a rectangle with width $f(x)$ and height $g(x)$. Its area is $A = f(x) \cdot g(x)$.

Now suppose $x$ increases by a tiny amount $\Delta x$. Both the width and the height change:

• The width becomes $f(x + \Delta x) \approx f(x) + f’(x)\Delta x$
• The height becomes $g(x + \Delta x) \approx g(x) + g’(x)\Delta x$

The new area is approximately: $$(f + f’\Delta x)(g + g’\Delta x) = fg + f’g\Delta x + fg’\Delta x + f’g’(\Delta x)^2$$

The change in area is: $$\Delta A \approx f’g\Delta x + fg’\Delta x + f’g’(\Delta x)^2$$

When we divide by $\Delta x$ to get the rate of change: $$\frac{\Delta A}{\Delta x} \approx f’g + fg’ + f’g’\Delta x$$

As $\Delta x \to 0$, that last term vanishes, leaving us with: $$\frac{dA}{dx} = f’g + fg’$$

This is the product rule! The rectangle gains area in two ways: a horizontal strip (when the width grows) and a vertical strip (when the height grows). The tiny corner piece where both grow simultaneously becomes negligible as $\Delta x$ shrinks to zero.

The Quotient Rule: Statement and Intuition

When dividing functions instead of multiplying them, we need a different rule:

$$\boxed{\left(\frac{f}{g}\right)’ = \frac{f’g - fg’}{g^2}}$$

Or in Leibniz notation:

$$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f’(x) \cdot g(x) - f(x) \cdot g’(x)}{[g(x)]^2}$$

Notice the crucial differences from the product rule:

1. There is a subtraction in the numerator (not addition)
2. We must divide by $g^2$
3. The order matters: it is $f’g - fg’$, not $fg’ - f’g$

Memory Tricks for the Quotient Rule

The quotient rule is notoriously easy to mix up. Here are some memory devices that generations of calculus students have found helpful:

“Lo d-hi minus hi d-lo, over lo-lo”

If you think of the fraction $\frac{f}{g}$ as “hi over lo” (high over low), then:

• “Lo d-hi” means: the low function times the derivative of the high = $g \cdot f'$
• “minus hi d-lo” means: minus the high function times the derivative of the low = $-f \cdot g'$
• “over lo-lo” means: all divided by the low function squared = $g^2$

This gives: $\frac{g \cdot f’ - f \cdot g’}{g^2} = \frac{f’g - fg’}{g^2}$

“Bottom times derivative of top, minus top times derivative of bottom, all over bottom squared”

This is more verbose but completely explicit. Whatever helps you remember!

Why the Quotient Rule Works

You can actually derive the quotient rule from the product rule. If $h = \frac{f}{g}$, then $f = h \cdot g$.

Applying the product rule to $f = hg$: $$f’ = h’g + hg’$$

Solving for $h’$: $$h’ = \frac{f’ - hg’}{g} = \frac{f’ - \frac{f}{g} \cdot g’}{g} = \frac{f’g - fg’}{g^2}$$

This is the quotient rule! So if you ever forget it during an exam, you can rederive it from the product rule.

When to Use Which Rule

Use the product rule when:

• You have two (or more) functions being multiplied
• Neither factor is a constant
• Simplifying first would be more complicated than using the rule

Use the quotient rule when:

• You have one function divided by another
• Both numerator and denominator depend on $x$
• You cannot easily simplify to avoid the fraction

Consider simplifying first when:

• One factor is a constant (just use the constant multiple rule)
• The expression can be rewritten without products or quotients
• Expansion or reduction would make differentiation easier

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Notation and Terminology

Term	Symbol/Formula	Meaning
Product rule	$(fg)’ = f’g + fg'$	Derivative of a product of two functions
Quotient rule	$\left(\frac{f}{g}\right)’ = \frac{f’g - fg’}{g^2}$	Derivative of a quotient of two functions
Leibniz notation (product)	$\frac{d}{dx}[f \cdot g] = \frac{df}{dx} \cdot g + f \cdot \frac{dg}{dx}$	Product rule in differential notation
Leibniz notation (quotient)	$\frac{d}{dx}\left[\frac{f}{g}\right] = \frac{\frac{df}{dx} \cdot g - f \cdot \frac{dg}{dx}}{g^2}$	Quotient rule in differential notation
Extended product rule	$(fgh)’ = f’gh + fg’h + fgh'$	Product rule for three functions
“Hi over lo”	$\frac{f}{g}$	Mnemonic: $f$ is “hi” (high), $g$ is “lo” (low)

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Examples

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Key Properties and Rules

The Product Rule in Multiple Forms

1. Prime notation: $(fg)’ = f’g + fg'$

2. Leibniz notation: $\frac{d}{dx}[fg] = \frac{df}{dx} \cdot g + f \cdot \frac{dg}{dx}$

3. Differential notation: $d(fg) = g , df + f , dg$

The Quotient Rule in Multiple Forms

1. Prime notation: $\left(\frac{f}{g}\right)’ = \frac{f’g - fg’}{g^2}$

2. Leibniz notation: $\frac{d}{dx}\left[\frac{f}{g}\right] = \frac{\frac{df}{dx} \cdot g - f \cdot \frac{dg}{dx}}{g^2}$

Extended Product Rule

For products of multiple functions:

• Two functions: $(fg)’ = f’g + fg'$
• Three functions: $(fgh)’ = f’gh + fg’h + fgh'$
• Four functions: $(fghk)’ = f’ghk + fg’hk + fgh’k + fghk'$

The pattern continues: each function takes a turn being differentiated while the others remain unchanged, and all these terms are added together.

Common Mistakes to Avoid

1. Thinking $(fg)’ = f’g’$ - This is the most common error. The product rule has addition, not a simple product of derivatives.

2. Forgetting the order in the quotient rule - Remember: it is $f’g - fg’$, not $fg’ - f’g$. The derivative of the top comes first in the first term.

3. Forgetting to square the denominator - The quotient rule has $g^2$ in the denominator, not just $g$.

4. Not simplifying first when possible - Sometimes $\frac{x^2}{x}$ is just $x$, and its derivative is simply 1. Do not make things harder than necessary.

5. Applying the wrong rule - Make sure you identify whether you have a product (use product rule) or a quotient (use quotient rule).

Special Cases

When one factor is a constant: If $c$ is a constant, then $(c \cdot f)’ = c \cdot f’$. You do not need the product rule because $c’ = 0$.

When the denominator is a constant: If $c$ is a constant, then $\left(\frac{f}{c}\right)’ = \frac{f’}{c}$. You do not need the quotient rule.

Quotient rule alternative: You can rewrite $\frac{f}{g} = f \cdot g^{-1}$ and use the product rule combined with the chain rule (covered in a later lesson). Some people prefer this approach.

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Real-World Applications

Revenue and Economics

In business, revenue often equals price times quantity: $R = P \times Q$. But here is the catch: if you are a large enough seller, the price you can charge depends on how much you sell. More supply might mean lower prices to attract buyers.

So if $P(q)$ is the price as a function of quantity, and your revenue is: $$R(q) = P(q) \cdot q$$

To find the marginal revenue (how much extra revenue you get from selling one more unit), you need the product rule: $$R’(q) = P’(q) \cdot q + P(q) \cdot 1 = P’(q) \cdot q + P(q)$$

This tells us that marginal revenue has two components:

1. The revenue from the additional unit sold at the current price: $P(q)$
2. The change in revenue from all units due to the price change: $P’(q) \cdot q$

Since $P’(q)$ is typically negative (higher quantity means lower price), the second term reduces the marginal revenue. This is why monopolists produce less than competitive markets!

Population Density

Population density is population divided by area: $$D = \frac{P}{A}$$

If a city’s population and area are both changing over time, how fast is the density changing?

Using the quotient rule with respect to time $t$: $$\frac{dD}{dt} = \frac{\frac{dP}{dt} \cdot A - P \cdot \frac{dA}{dt}}{A^2}$$

This formula reveals that density increases when population grows faster than area expands, and decreases when area grows faster than population.

Chemical Concentration

In chemistry, concentration equals amount of substance divided by volume: $$C = \frac{n}{V}$$

If you are adding a solute to a solution while also diluting it (adding solvent), both $n$ and $V$ change. The rate of concentration change is: $$\frac{dC}{dt} = \frac{\frac{dn}{dt} \cdot V - n \cdot \frac{dV}{dt}}{V^2}$$

This is crucial in pharmacology, where drug concentrations in the body change as drugs are metabolized and fluids are processed.

Physics: Momentum

In physics, momentum equals mass times velocity: $p = mv$. For objects with changing mass (like rockets burning fuel), both $m$ and $v$ change with time. The force is: $$F = \frac{dp}{dt} = \frac{d}{dt}[mv] = m\frac{dv}{dt} + v\frac{dm}{dt}$$

The first term is the familiar $ma$ from Newton’s second law. The second term accounts for the force due to changing mass, which is essential for understanding rocket propulsion.

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Self-Test Problems

Test your understanding with these practice problems. Try each one on paper before revealing the answer.

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Summary

Here are the essential takeaways from this lesson:

• The product rule states that $(fg)’ = f’g + fg’$. When two functions are multiplied, each takes a turn being differentiated while the other remains unchanged, and the results are added.

• The rectangle analogy explains why: a growing rectangle gains area through a horizontal strip (from width growth) and a vertical strip (from height growth). The corner piece becomes negligible.

• The quotient rule states that $\left(\frac{f}{g}\right)’ = \frac{f’g - fg’}{g^2}$. Remember “lo d-hi minus hi d-lo, over lo-lo” to keep the order straight.

• The order matters in the quotient rule: it is $f’g - fg’$, and switching the order gives the wrong sign.

• Common mistake warning: $(fg)’ \neq f’g’$. This false rule would give wrong answers, as we verified with $x \cdot x = x^2$.

• Simplify when possible. Sometimes you can avoid using these rules entirely by simplifying the expression first.

• The extended product rule handles products of three or more functions: each factor takes its turn being differentiated while the others remain unchanged.

• Real-world applications include revenue analysis (price times quantity), population density (population over area), chemical concentrations, and momentum in physics.

• Practice is essential. These rules become automatic with repetition, so work through plenty of examples until the mechanics feel natural.

With the power and product rules mastered, you can now differentiate any polynomial and many rational functions. The chain rule, which handles compositions of functions, will expand your capabilities even further.