The product rule is one of the most important rules in differential calculus. It is often introduced as a formula to memorize:

(fg)’=f’g+fg’

But there is a much deeper story behind this formula.

Why are there two terms?

Why isn’t the derivative simply f’g’?

And why do the original functions f and g appear alongside their derivatives?

The answer becomes surprisingly intuitive when we think about the area of a rectangle whose width and height are both changing.

1. Start with a rectangle

Imagine a rectangle whose:

width is f(x)
height is g(x)

Its area is therefore:

$A(x)=f(x)g(x)$

Now increase x by a very small amount $\Delta x$ .

Both dimensions may change.

The width changes by $\Delta f$ , while the height changes by $\Delta g$ .

So the new dimensions are:

$f+\Delta f$

and

$g+\Delta g$

The new area is therefore:

$(f+\Delta f)(g+\Delta g)$

This simple rectangle gives us almost the entire intuition behind the product rule.

2. Expand the new area

Using ordinary algebra:

$(f+\Delta f)(g+\Delta g)=fg+g\Delta f+f\Delta g+\Delta f\Delta g$

The original area was:

$fg$

Therefore, the change in area is:

$\Delta A=g\Delta f+f\Delta g+\Delta f\Delta g$

This equation is the key.

The change in the product consists of three pieces.

3. Where do the three pieces come from?

The first strip

Suppose the width changes by $\Delta f$ while the old height remains $g$ .

The resulting additional area is:

$g\Delta f$

This represents the change caused by the first factor.

The second strip

Now consider the height changing by $\Delta g$ while the old width remains $f$ .

The additional area is:

$f\Delta g$

This represents the change caused by the second factor.

The tiny corner

There is also a small corner where both dimensions have changed.

Its area is:

$\Delta f\Delta g$

Therefore:

$\Delta A=g\Delta f+f\Delta g+\Delta f\Delta g$

This is the geometric meaning of the algebraic expansion.

4. Turn change into rate of change

A derivative measures the rate at which something changes.

So divide the entire equation by $\Delta x$ :

$\frac{\Delta A}{\Delta x}=g\frac{\Delta f}{\Delta x}+f\frac{\Delta g}{\Delta x}+\frac{\Delta f\Delta g}{\Delta x}$

Now we make $\Delta x$ smaller and smaller.

In the limit:

$\lim_{\Delta x\to0}\frac{\Delta f}{\Delta x}=f'(x)$

and:

$\lim_{\Delta x\to0}\frac{\Delta g}{\Delta x}=g'(x)$

So the first two terms become:

$gf'(x)$

and:

$fg'(x)$

But what happens to the tiny corner?

5. Why does the tiny corner disappear?

As $\Delta x$ becomes very small, the changes in the two functions also become small.

Approximately:

$\Delta f\approx f'\Delta x$

and:

$\Delta g\approx g'\Delta x$

Therefore:

$\Delta f\Delta g\approx f'g'(\Delta x)^2$

Now divide by $\Delta x$ :

$\frac{\Delta f\Delta g}{\Delta x}\approx f'g'\Delta x$

As $\Delta x\to0$ :

$f'g'\Delta x\to0$

So the corner becomes negligible.

This is an important idea in calculus:

A product of two tiny changes is of a smaller order than either individual change.

Therefore, only the two first-order contributions survive.

6. We have discovered the product rule

We are left with:

$A'(x)=g(x)f'(x)+f(x)g'(x)$

Rearranging the terms:

$\boxed{(fg)'=f'g+fg'}$

This is the product rule.

It isn’t an arbitrary formula.

It comes directly from the geometry of a changing rectangle.

7. A rigorous proof from the definition of the derivative

The geometric argument gives us the intuition. Now let’s prove the same result directly from the definition of a derivative.

Let:

$h(x)=f(x)g(x)$

By definition:

$h'(x)=\lim_{\Delta x\to0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}$

The difficulty is that both functions are changing simultaneously.

We need to separate their contributions.

We do this by adding and subtracting the intermediate quantity:

$f(x+\Delta x)g(x)$

So the numerator becomes:

$f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x)+f(x+\Delta x)g(x)-f(x)g(x)$

Now group the terms:

$h'(x)=\lim_{\Delta x\to0}\left[f(x+\Delta x)\frac{g(x+\Delta x)-g(x)}{\Delta x}+g(x)\frac{f(x+\Delta x)-f(x)}{\Delta x}\right]$

Now take the limit.

As $\Delta x\to0$ :

$f(x+\Delta x)\to f(x)$

and:

$\frac{g(x+\Delta x)-g(x)}{\Delta x}\to g'(x)$

Similarly:

$\frac{f(x+\Delta x)-f(x)}{\Delta x}\to f'(x)$

Therefore:

$h'(x)=f(x)g'(x)+g(x)f'(x)$

Hence:

$\boxed{(fg)'=f'g+fg'}$

The product rule is proven.

8. Why isn’t the answer $f'g'$ ?

This is probably the most common intuitive objection.

Suppose:

$A=fg$

The area can change because the width changes.

That contribution is approximately:

$g\Delta f$

The area can also change because the height changes.

That contribution is approximately:

$f\Delta g$

Therefore:

$\Delta A\approx g\Delta f+f\Delta g$

The two contributions are added, not multiplied.

That’s why the derivative contains:

$f'g+fg'$

rather than:

$f'g'$ .

9. Why is there a plus sign?

The plus sign is now easy to understand.

The product can change in two different ways:

First factor changes:

$f'g$

Second factor changes:

$fg'$

The total change is the combination of both effects:

$\boxed{f'g+fg'}$

So the product rule can be remembered conceptually as:

Change caused by the first factor + change caused by the second factor.

10. The deeper meaning of the two terms

Consider:

$A=fg$

The first term:

$f'g$

means:

Let $g$ temporarily behave as though it were constant. How much does the product change because $f$ changes?

The second term:

$fg'$

means:

Let $f$ temporarily behave as though it were constant. How much does the product change because $g$ changes?

Then we add the two effects.

That is the fundamental idea behind the product rule.

11. A simple example

Take:

$f(x)=x^2$

and:

$g(x)=x$

Their product is:

$f(x)g(x)=x^3$

We know directly that:

$\frac{d}{dx}x^3=3x^2$

Now let’s use the product rule.

First:

$f'(x)=2x$

and:

$g'(x)=1$

Therefore:

$(x^2\cdot x)'=(2x)(x)+(x^2)(1)$

So:

$=2x^2+x^2$

and finally:

$=3x^2$

Exactly what we expected.

12. A real-world interpretation

The product rule isn’t limited to geometry.

Suppose some quantity is defined as:

$Q(t)=m(t)v(t)$

where both mass $m$ and velocity $v$ can change over time.

Then:

$Q'(t)=m'(t)v(t)+m(t)v'(t)$

There are two sources of change:

The quantity changes because the mass changes.
The quantity changes because the velocity changes.

Again, the total rate of change is the sum of the two effects.

This same mathematical structure appears throughout physics, engineering, economics, biology and other quantitative fields.

13. The beautiful connection with ordinary algebra

There is a particularly elegant way to see the product rule.

Start with the ordinary algebraic identity:

$(f+\Delta f)(g+\Delta g)=fg+g\Delta f+f\Delta g+\Delta f\Delta g$

Subtract the original product:

$\Delta(fg)=g\Delta f+f\Delta g+\Delta f\Delta g$

Now imagine that the changes become infinitesimally small.

The final term becomes negligible:

$\Delta f\Delta g\to0$

Therefore:

$d(fg)=g,df+f,dg$

Divide by $dx$ :

$\frac{d(fg)}{dx}=g\frac{df}{dx}+f\frac{dg}{dx}$

And therefore:

$\boxed{(fg)'=f'g+fg'}$

So calculus has not abandoned ordinary algebra.

The product rule is essentially the algebra of a changing product viewed at an infinitesimally small scale.

14. The most intuitive way to remember it

Instead of memorizing:

f’g+fg’

as a mysterious formula, think:

A product changes because either factor can change.

So:

$\text{total change}=\text{change from }f+\text{change from }g$

The change from $f$ is:

$f'g$

The change from $g$ is:

$fg'$

Therefore:

$\boxed{(fg)'=f'g+fg'}$

15. The big picture

The product rule is a beautiful example of how calculus emerges naturally from simple algebra.

Start with:

$(f+\Delta f)(g+\Delta g)$

Expand:

$=fg+g\Delta f+f\Delta g+\Delta f\Delta g$

Subtract the original product:

$\Delta(fg)=g\Delta f+f\Delta g+\Delta f\Delta g$

Divide by $\Delta x$ :

$\frac{\Delta(fg)}{\Delta x}=g\frac{\Delta f}{\Delta x}+f\frac{\Delta g}{\Delta x}+\frac{\Delta f\Delta g}{\Delta x}$

Take the limit:

$\boxed{\frac{d}{dx}(fg)=f'g+fg'}$

The tiny corner disappears.

The two first-order changes remain.

And the mysterious product rule emerges naturally.

Final takeaway

The product rule is not something that needs to be accepted on faith.

It follows from a simple observation:

When two quantities multiply, the product can change because the first quantity changes, because the second quantity changes, or because both change simultaneously.

The simultaneous-change term is second-order and disappears in the infinitesimal limit.

What remains are the two first-order effects:

$\boxed{\text{first factor changes}+\text{second factor changes}}$

which gives:

$\boxed{(fg)'=f'g+fg'}$

Once you see the changing rectangle, the product rule stops looking like a formula to memorize and starts looking like something that had to be true.

The Product Rule: A Mathematical Proof and the Intuition Behind It

1. Start with a rectangle

2. Expand the new area

3. Where do the three pieces come from?

The first strip

The second strip

The tiny corner

4. Turn change into rate of change

5. Why does the tiny corner disappear?

6. We have discovered the product rule

7. A rigorous proof from the definition of the derivative

8. Why isn’t the answer $f'g'$ ?

9. Why is there a plus sign?

10. The deeper meaning of the two terms

11. A simple example

12. A real-world interpretation

13. The beautiful connection with ordinary algebra

14. The most intuitive way to remember it

15. The big picture

Final takeaway

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DavidsonNext: AP® Calculus: Challenging Concepts from Calculus AB & Calculus BC

Calculus 1A: Differentiation

Calculus 1B: Integration

Calculus 1C: Coordinate Systems & Infinite Series

1. Start with a rectangle

2. Expand the new area

3. Where do the three pieces come from?

The first strip

The second strip

The tiny corner

4. Turn change into rate of change

5. Why does the tiny corner disappear?

6. We have discovered the product rule

7. A rigorous proof from the definition of the derivative

8. Why isn’t the answer ?

9. Why is there a plus sign?

10. The deeper meaning of the two terms

11. A simple example

12. A real-world interpretation

13. The beautiful connection with ordinary algebra

14. The most intuitive way to remember it

15. The big picture

Final takeaway

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Like this:

DavidsonNext: AP® Calculus: Challenging Concepts from Calculus AB & Calculus BC

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Calculus 1A: Differentiation

Calculus 1B: Integration

Calculus 1C: Coordinate Systems & Infinite Series

8. Why isn’t the answer $f'g'$ ?