Introduction

At the end of the previous post I promised to look next at why you cannot just differentiate each factor separately when differentiating a product, or a quotient, of two functions. That is today's story.

Let us first put a very natural misconception on the table. When we differentiate the product \(f(x)\,g(x)\) of two functions \(f\) and \(g\), the mind whispers:

"Can't I just differentiate each and multiply? So isn't \((fg)' = f'g'\)?"

It sounds plausible. But it is wrong. The correct answer has two terms.

$$(fg)' = f'g + fg'$$

Why two and not one? This post builds the answer from a single picture: the area of a rectangle. Once the picture stands, why the derivative of the quotient \(f/g\) suddenly carries a \(g^2\) and a minus sign flows out of that same picture — with nothing to memorize.

1. The product rule — the area grows in two directions

Think of the product \(f(x)\,g(x)\) as the area of a rectangle. If the width is \(f\) and the height is \(g\), then its area is exactly \(fg\). Now nudge \(x\) by a tiny \(\Delta x\), and both the width and the height grow a little.

  • The width \(f\) grows by \(\Delta f\) (its growth rate is \(f'\), so \(\Delta f \approx f'\,\Delta x\)).
  • The height \(g\) grows by \(\Delta g\) (likewise \(\Delta g \approx g'\,\Delta x\)).

The rectangle is now a bit bigger. But looking at where the new area got attached, the places split into exactly three.

  • The right strip — a vertical strip whose height is still the original \(g\) but whose width is the new \(\Delta f\). Area \(= g \cdot \Delta f\).
  • The top strip — a horizontal strip whose width is still the original \(f\) but whose height rose by \(\Delta g\). Area \(= f \cdot \Delta g\).
  • The top-right corner — a tiny square that appears only where both width and height grew at once. Area \(= \Delta f \cdot \Delta g\).

The total added area is the sum of these three.

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

Differentiation is dividing this added area by \(\Delta x\) and sending \(\Delta x \to 0\). Divide.

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

In the first two terms, \(\dfrac{\Delta f}{\Delta x}\) approaches \(f'\) and \(\dfrac{\Delta g}{\Delta x}\) approaches \(g'\). The trouble is the last, corner term. Since \(\Delta f \approx f'\,\Delta x\), the corner area is \(\Delta f\,\Delta g \approx f'g'\,(\Delta x)^2\) — a size in which \(\Delta x\) is multiplied twice. Divide it by a single \(\Delta x\) and \(f'g'\,\Delta x\) remains, which as \(\Delta x \to 0\) vanishes entirely.

  • Intuition: the corner appears only when 'both width and height grow.' Being a product of two small changes, it is a small thing times another small thing — an area negligible enough to drop.
  • Explanation: the right strip and the top strip are proportional to \(\Delta x\) once, so they survive; only the corner is proportional to \(\Delta x\) twice and gets erased in the limit.
  • Formula: so only the two strips survive.
$$(fg)' = f'g + fg'$$

Now you can feel why there are two terms. The area grows in two directions — rightward (because the width grew) and upward (because the height grew). One is "\(f\) changes while \(g\) stays still," the other is "\(g\) changes while \(f\) stays still" — and the product rule is the sum of those two shares.

2. So why is \(f'g'\) wrong — once, with numbers

Having seen it as a picture, let us confirm it with numbers. For the simplest example, take \(f(x) = x\) and \(g(x) = x\). Their product is \(f g = x \cdot x = x^2\), and we already know from the definition that \((x^2)' = 2x\).

Now set the two methods side by side.

MethodComputationResult
Wrong: \(f'g'\)\(1 \cdot 1\)\(1\)
Product rule: \(f'g + fg'\)\(1 \cdot x + x \cdot 1\)\(2x\)

Since \(f' = 1\) and \(g' = 1\), \(f'g'\) comes out as the constant \(1\). But the true answer is \(2x\). Only the product rule, summing the two terms, hits \(2x\) exactly. Differentiating each and multiplying counts only one of the two strips and drops the other entirely, which is why the answer misses.

3. The quotient rule — rewrite division as "flip and multiply"

Now for division. Differentiating \(\dfrac{f}{g}\) is usually memorized like this.

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

There is a \(g^2\) in the denominator and a minus on top. Where did these two come from? Instead of memorizing a new formula, rewrite the division as multiplication and ride the product rule we just built. Dividing is multiplying by the reciprocal.

$$\frac{f}{g} = f \cdot \frac{1}{g}$$

Now we can use the product rule — but first we need the derivative \(\left(\dfrac{1}{g}\right)'\). This can be found by the chain rule or from the definition, and the result is the key.

$$\left(\frac{1}{g}\right)' = -\frac{g'}{g^2}$$
  • Intuition: when \(g\) grows, its reciprocal \(1/g\) shrinks. So the change of \(1/g\) runs opposite to the growth of \(g\) — the minus is born right here.
  • Explanation: why \(g^2\)? Differentiating \(1/g\) from the definition, combining fractions gives a denominator \(g \cdot (g+\Delta g)\), which in the limit hardens into \(g \cdot g = g^2\). Differentiating a reciprocal makes the denominator swell into a square.

Holding this \(\left(\dfrac{1}{g}\right)' = -\dfrac{g'}{g^2}\), apply the product rule to \(f \cdot \dfrac{1}{g}\).

$$\left(f \cdot \frac{1}{g}\right)' = f' \cdot \frac{1}{g} + f \cdot \left(\frac{1}{g}\right)' = \frac{f'}{g} + f \cdot \left(-\frac{g'}{g^2}\right)$$

Tidying up, this becomes:

$$= \frac{f'}{g} - \frac{fg'}{g^2}$$

Putting the first term's denominator \(g\) over \(g^2\) (multiply top and bottom by \(g\) to get \(\dfrac{f'g}{g^2}\)) gathers the two fractions under one roof.

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

The memorized formula has reproduced itself. But now its whole appearance is explained.

  • The minus — comes from the reciprocal property that \(1/g\) shrinks as \(g\) grows.
  • The \(g^2\) — comes from the square the denominator swells into when the reciprocal \(1/g\) is differentiated.
  • The two terms \(f'g\) and \(fg'\) — are, in the end, descendants of those two strips from the product rule.

The quotient rule is not a separate rule dropped from the sky; it is the product rule with one part — 'differentiate the reciprocal' — plugged in.

4. Check it yourself in the interactive below

Grow the rectangle with the sliders and follow, by eye, the two strips that grow and the corner that vanishes.

  • Switch between Product / Quotient mode with the buttons. Product mode is a rectangle of width \(f\) and height \(g\); quotient mode has height \(1/g\).
  • Use the width \(f\) and height \(g\) sliders to set the rectangle's size. (The growth rates \(f'\) and \(g'\) are shown on screen as fixed values.)
  • Shrink the gap \(\Delta x\) slider toward zero and watch the green strip (right) and blue strip (top) survive while only the red corner shrinks away rapidly. This is the scene where the corner term in \((fg)'\) is erased in the limit.
  • In quotient mode, watch the top strip grow downward (in the minus direction). The height \(1/g\) shrinks opposite to \(g\), and this is the identity of the minus sign attached to the quotient rule.
The product rule and the quotient rule — two strips of a rectangle
In product mode, shrinking the gap slider keeps the green and blue strips while only the red corner vanishes. Switch to quotient mode and the top strip grows downward, showing where the minus sign comes from.

Handling it, you feel with your fingertips that both the two terms of the product rule and the square-and-minus of the quotient rule come, in the end, from one and the same rectangle.

5. Summary — two rules out of a single rectangle

  • Product rule: seeing \(fg\) as a rectangle's area, the area grows in two directions — the right strip \(g\,\Delta f\) and the top strip \(f\,\Delta g\). The corner \(\Delta f\,\Delta g\) is proportional to \(\Delta x\) twice and vanishes in the limit. Hence \((fg)' = f'g + fg'\).
  • Why not \(f'g'\): differentiating each and multiplying counts only one of the two strips, so the answer misses (confirmed with \(x \cdot x\)).
  • Quotient rule: rewrite \(f/g\) as \(f \cdot (1/g)\), plug the reciprocal derivative \(\left(1/g\right)' = -g'/g^2\) into the product rule, and \(\left(f/g\right)' = \dfrac{f'g - fg'}{g^2}\) falls out on its own. The minus comes from the reciprocal, the \(g^2\) from the square of the reciprocal's derivative.

The two rules look like different formulas, but their root is a single rectangle whose width and height grow together. The product rule is that root, and the quotient rule extends one more branch — the reciprocal — from it.

Closing

By today we have secured the rules for differentiating products and quotients. But one big part still remains. Why differentiating \(\sin x\) makes \(\cos x\) pop out, and how to differentiate the layered structure of a function inside a function — the chain rule and the derivatives of trigonometric functions are the next story. Just as we built the product rule from a rectangle today, we will see those grow out of a single picture too.