The product rule

If $y = uv$ , where $u$ and $v$ are both functions of $x$ , then

\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}.

In words: differentiate the first and keep the second, then keep the first and differentiate the second, and add the two results.

Before you start

You need two facts.

The derivative is a limit. The derivative measures a small change in $y$ against the small change in $x$ that caused it: $ $\frac{dy}{dx} = \lim_{\delta x \to 0}\frac{\delta y}{\delta x}.$ $
A sum that tends to a limit. If one part of an expression tends to a limit and another part tends to $0$ , the second part disappears in the limit. That is what decides the fate of the corner in this proof.

Predict first

Picture a rectangle with width $u$ and height $v$ , so its area is $uv$ . Now let both sides grow: the width grows by $\delta u$ and the height grows by $\delta v$ .

Before you read on, predict: is the growth in area the two new strips alone — the strip along the right and the strip along the top — or is something else there as well? Draw the grown rectangle and look at the corner where the two strips meet.

Now test it. Move each slider and watch the two strips and the corner. Then explain, in your own words, why the corner shrinks faster than either strip as the two growths get smaller.

The proof

The symbol $\delta$ (delta) stands for a small change. Let $x$ change by $\delta x$ . Because $u$ and $v$ both depend on $x$ , that makes $u$ change by some $\delta u$ and $v$ change by some $\delta v$ . The changes are linked to $\delta x$ : as $\delta x$ tends to $0$ , so do $\delta u$ and $\delta v$ .

The new area

The old rectangle has area

uv.

The new rectangle has width $u + \delta u$ and height $v + \delta v$ , so its area is the product

(u + \delta u)(v + \delta v).

Multiply out the brackets, taking every term by every term:

(u + \delta u)(v + \delta v) = uv + u\,\delta v + v\,\delta u + \delta u\,\delta v.

You can read the four pieces in the figure:

$uv$ is the original rectangle;
$v\,\delta u$ is the strip added along the right;
$u\,\delta v$ is the strip added along the top;
$\delta u\,\delta v$ is the small corner where those strips meet.

The change in area

Subtract the old area $uv$ :

\delta(uv) = (u + \delta u)(v + \delta v) - uv = u\,\delta v + v\,\delta u + \delta u\,\delta v.

Both strips are there, and so is the corner. The prediction asked whether the two strips were the whole change. They are not, and the corner is the missing piece.

Divide by the change in x

Now do what the definition of the derivative asks: divide the change in $uv$ by the change in $x$ .

\frac{\delta(uv)}{\delta x} = \frac{u\,\delta v + v\,\delta u + \delta u\,\delta v}{\delta x} = u\frac{\delta v}{\delta x} + v\frac{\delta u}{\delta x} + \delta u\frac{\delta v}{\delta x}.

Look closely at the third term. Dividing $\delta u\,\delta v$ by $\delta x$ gives the $\delta x$ to the $\delta v$ , which leaves

\delta u \times \frac{\delta v}{\delta x}.

That is the corner, still divided by $\delta x$ .

Take the limit

Let $\delta x \to 0$ . Each piece does something simple.

$u$ does not depend on $\delta x$ , so it stays $u$ ;
$\dfrac{\delta v}{\delta x}$ tends to the derivative $\dfrac{dv}{dx}$ ;
$v$ stays $v$ , and $\dfrac{\delta u}{\delta x}$ tends to $\dfrac{du}{dx}$ ;
$\delta u$ tends to $0$ .

Putting those together,

\frac{d(uv)}{dx} = \lim_{\delta x \to 0}\left(u\frac{\delta v}{\delta x} + v\frac{\delta u}{\delta x} + \delta u\frac{\delta v}{\delta x}\right) = u\frac{dv}{dx} + v\frac{du}{dx} + 0 \times \frac{dv}{dx}.

The third term is $0 \times \dfrac{dv}{dx}$ , and that is $0$ as long as $\dfrac{dv}{dx}$ is a finite number, which it is for every function met in this course. So

\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}.

That is the product rule.

Why the corner contributes nothing

This is the step that is usually skipped, so it is worth saying plainly. Compare the three terms that are left after the division by $\delta x$ .

The strip $v\,\delta u$ was one small amount, $\delta u$ , multiplied by $v$ . Dividing by $\delta x$ turns $\dfrac{\delta u}{\delta x}$ into the derivative $\dfrac{du}{dx}$ . Something survives, because one small length was divided by one small length.
The other strip behaves the same way, with $u$ and $v$ swapped.
The corner was two small amounts, $\delta u$ and $\delta v$ , multiplied together. The division by $\delta x$ removes one of them. A single factor of $\delta u$ is left behind, and $\delta u$ tends to $0$ .

Two small lengths cannot outrun one division by a small length, so the corner tends to $0$ while the strips tend to the two derivatives.

If that is hard to see, watch it in the figure. Shrink the two growths and the corner takes a smaller and smaller share of the whole change. In the limit the corner is nothing, and the two strips are everything.

Check your understanding

A rectangle is $8\text{ cm}$ wide and $5\text{ cm}$ tall. The width grows by $0.2\text{ cm}$ and the height grows by $0.1\text{ cm}$ . Work out the two strips and the corner, and check that the three pieces add to the exact change in area.
Where does the corner term $\delta u\,\delta v$ appear in $\dfrac{\delta(uv)}{\delta x}$ after the division by $\delta x$ ?
Use the product rule with $u = x$ and $v = x$ to differentiate $x^2$ . What does the corner term look like here, and why does it vanish in the limit?

Answers

The right strip is $5 \times 0.2 = 1\text{ cm}^2$ , the top strip is $8 \times 0.1 = 0.8\text{ cm}^2$ , and the corner is $0.2 \times 0.1 = 0.02\text{ cm}^2$ . They add to $1.82\text{ cm}^2$ . The areas are $40$ and $8.2 \times 5.1 = 41.82$ , and $41.82 - 40 = 1.82$ , so the three pieces are the whole change.
It becomes $\delta u\,\dfrac{\delta v}{\delta x}$ , the corner still divided by $\delta x$ . As $\delta x \to 0$ , the factor $\delta u$ tends to $0$ , so the whole term tends to $0$ .
With $u = v = x$ , the product rule gives $x \times 1 + x \times 1 = 2x$ . The corner is $\delta x \times \delta x = (\delta x)^2$ , which divided by $\delta x$ is $\delta x$ , and that tends to $0$ .

A common error

Do not differentiate a product by differentiating each part and multiplying. The derivative of $uv$ is not $\dfrac{du}{dx} \times \dfrac{dv}{dx}$ . Check it on the simplest case, $u = v = x$ : multiplying the derivatives gives $1 \times 1 = 1$ , but the derivative of $x^2$ is $2x$ . The rule adds the two strips together. The corner is what makes the grown rectangle exactly a rectangle, and the corner is what disappears.