Product and quotient rules

Why a product needs its own rule

The derivative of a product is not the product of the derivatives. Lesson 2.3.1 showed one case: $x^2 \cdot x^3 = x^5$ has slope $5x^4$. Multiplying the two slopes gives $2x \cdot 3x^2 = 6x^3$ instead, which is 6 at $x = 1$ where the true slope is 5.

When both factors change, the product changes for two reasons at once. A picture shows how the two combine.

The area picture

Think of two functions $f(x)$ and $g(x)$ as the width and height of a rectangle. Its area is the product $f \cdot g$. When $x$ grows by $h$, the width grows by $\Delta f$ and the height by $\Delta g$.

Take $f(x) = x^2$ and $g(x) = x + 2$ at $x = 3$. The rectangle is 9 wide and 5 tall, with area 45. A step of $h = 0.4$ makes the width 11.56 and the height 5.4.

The new area is the old area plus three pieces.

• A strip $g \cdot \Delta f$, from the width growing.
• A strip $f \cdot \Delta g$, from the height growing.
• A corner $\Delta f \cdot \Delta g$, from both growing at once.

Now divide each piece by $h$ and let $h$ shrink.

The first row tends to $g \cdot f' = 5 \times 6 = 30$. The second is $f \cdot g' = 9 \times 1 = 9$. The corner is a small amount times a small amount, so even after dividing by $h$ it vanishes.

So the slope of $x^2(x + 2)$ at 3 is $30 + 9 = 39$. Algebra agrees: $x^2(x + 2) = x^3 + 2x^2$, with derivative $3x^2 + 4x$, which is $27 + 12 = 39$ at 3. Multiplying the slopes would give $6 \times 1 = 6$.

The units work like area. If the width and height were in metres and $x$ in seconds, each strip would grow in square metres per second.

A second example

For $x^2(x^3 + 1)$, take $f = x^2$ and $g = x^3 + 1$, so $f' = 2x$ and $g' = 3x^2$. The product rule gives $2x(x^3 + 1) + x^2 \cdot 3x^2 = 5x^4 + 2x$. At $x = 1$ the slope is 7.

Expanding first gives the same: $x^5 + x^2$ has derivative $5x^4 + 2x$. The product rule matters when expanding is long or impossible, as with the exponentials of lesson 2.3.4.

When the factors pull in opposite directions

In $x(6 - x)$ the first factor rises with slope 1 and the second falls with slope $-1$. The product rule gives $1 \cdot (6 - x) + x \cdot (-1) = 6 - 2x$. The two terms have opposite signs, and the larger one wins.

At 1 the rising factor wins and the product climbs. At 3 the terms cancel and the product is flat for an instant. At 5 the falling factor wins.

Constant factors and three factors

If one factor is a constant $c$, its slope is 0 and the rule gives $0 \cdot g + c\,g' = c\,g'$. That is the constant rule of lesson 2.3.1, so nothing changes.

With three factors, apply the rule twice. Each term differentiates one factor and keeps the other two: $(fgk)' = f'gk + fg'k + fgk'$. For $x \cdot x \cdot x$ that gives $x^2 + x^2 + x^2 = 3x^2$, as the power rule says.

The quotient rule

A quotient $\frac{f}{g}$ has its own rule. It looks like the product rule with a minus sign and a new bottom.

Take $\frac{x}{x + 1}$, with $f = x$ and $g = x + 1$, so $f' = 1$ and $g' = 1$. The rule gives

At $x = 1$ that is $\frac{1}{4}$. The central difference from lesson 2.2.2 agrees.

The rule also reproduces a result from lesson 2.2.2. For $\frac{1}{x}$, take $f = 1$ and $g = x$: the slope is $\frac{0 \cdot x - 1 \cdot 1}{x^2} = -\frac{1}{x^2}$.

Signs, and where a quotient breaks

The bottom $g^2$ is never negative, so the sign of the slope is the sign of the top, $f'g - fg'$. Swapping the two terms flips the sign: it would give $-\frac{1}{4}$ for $\frac{x}{x + 1}$ at 1, which is wrong.

For $\frac{x^2}{x^2 + 1}$ the top simplifies to $2x$, so the slope is $\frac{2x}{(x^2 + 1)^2}$. It is $-0.5$ at $x = -1$, 0 at $x = 0$ and $0.5$ at $x = 1$: falling, flat, then rising.

Where the bottom is 0, the quotient has no value, so it has no slope. $\frac{x}{x + 1}$ has none at $x = -1$.

Use the quickest rule

The quotient rule always works when the bottom is not 0, but it is not always the quickest.

• Constant top: $\frac{3}{x^2} = 3x^{-2}$, so the power rule gives $-\frac{6}{x^3}$.
• Constant bottom: $\frac{x^2 + 1}{4}$ is $\frac{1}{4}(x^2 + 1)$, with slope $\frac{x}{2}$.
• A single power at the bottom: split it, as in lesson 2.3.1.
• A sum at the bottom: use the quotient rule. $\frac{1}{x^2 + 1}$ has slope $\frac{0 - 1 \cdot 2x}{(x^2 + 1)^2} = -\frac{2x}{(x^2 + 1)^2}$.

Lesson 2.3.3 handles a function inside another function, and lesson 2.6.2 runs the product rule backwards as integration by parts. Case study 2.3.7 applies the product rule to revenue, a price times a quantity.

Common mistakes

1. $(fg)' \neq f'g'$. For $x^2(x + 2)$ at 3 that gives 6; the product rule gives 39.
2. Each term differentiates exactly one factor. $f'g + fg'$, never $f'g'$ or $f'g + f'g$.
3. Keep the order on top of the quotient rule: $f'g - fg'$. Swapping flips the sign of the answer.
4. Square the bottom. The quotient rule divides by $g^2$, not by $g$.
5. No slope where the bottom is 0. $\frac{x}{x + 1}$ has none at $-1$.
6. Same names, different rules. The product and quotient rules of exponents (lesson 1.4.1) and logs (lesson 1.4.3) combine powers; these rules find slopes.