The Product Rule

A shop's revenue is price times the number it sells; drop the price in a sale and sales climb, so both factors move at once — is revenue rising or falling? Any quantity that is one changing thing multiplied by another — an area as both sides grow, electrical power as voltage and current shift — raises the same question, and the product rule answers it.

You have a rule for sums: differentiate term by term, and the pieces never interfere — (u + v)' = u' + v'. It works so smoothly that when a product of two functions turns up, like x^2 \sin x or (x^2+1)(x^3-2), almost everyone writes down the same guess without a moment's hesitation:

\frac{d}{dx}\big[u\,v\big] \overset{?}{=} u'\,v' \qquad \text{(the guess)}

Differentiate each factor, multiply the answers. It looks right. It is wrong — and you can demolish it in one line. Test it on the simplest product in the world, x \cdot x. The guess says the derivative is 1 \cdot 1 = 1, a constant. But x \cdot x = x^2, whose derivative you already know is 2x — not constant at all. The guess isn't slightly off; it's a completely different function. One counterexample, and the "obvious" rule is dead.

The correct rule is stranger and more interesting: the slope of a product is not the product of the slopes, because a product grows on two fronts at once. Each factor takes a turn being the one that changes, while the other holds still — and you add the two contributions. This page shows you the rule, shows you why it must be true (a growing rectangle tells the whole story), and puts it to work.

The rule itself

If u(x) and v(x) are both differentiable at x, then so is their product, and \frac{d}{dx}\big[u\,v\big] = u'\,v + u\,v'.
In Leibniz notation: d(uv) = v\,du + u\,dv — a tiny change in the product is the second factor times the change in the first, plus the first factor times the change in the second.

Read it as a chant: differentiate the first, leave the second; leave the first, differentiate the second — and add. Exactly one factor wears the prime in each term. Never both, never neither.

Sanity-check it on the product that killed the guess. With u = x and v = x:

\frac{d}{dx}\big[x \cdot x\big] = (1)(x) + (x)(1) = 2x \; \checkmark

Precisely the 2x the power rule demands. The two terms are the two fronts on which the product grows — and the picture below shows exactly what those two fronts look like.

Why: a growing rectangle

Here is the heart of the page. Picture u and v as the two sides of a rectangle, so the product uv is its area. As x ticks forward, both sides grow a little: u grows by du and v by dv. How much new area appears?

Look at where it appears. A thin strip along the top: width u, thickness dv, area u\,dv. A thin strip up the side: height v, thickness du, area v\,du. And one tiny corner square, du \cdot dv — a small number times a small number, far tinier than either strip. The algebra says the same thing all at once:

(u + du)(v + dv) = uv + u\,dv + v\,du + \underbrace{du\,dv}_{\text{negligible}}

Try it with honest numbers. Take a 3 \times 2 rectangle (area 6) and grow both sides by 0.1. The new area is 3.1 \times 2.1 = 6.51, so the growth is 0.51. Where did it come from? Top strip 3 \times 0.1 = 0.3, side strip 2 \times 0.1 = 0.2, corner 0.1 \times 0.1 = 0.01. The strips carry 0.5 of the growth; the corner contributes a rounding error — and it only gets more negligible as the growth shrinks. Drag the slider to grow the sides and watch the two strips (and the shrinking corner) appear:

Now turn "growth" into "rate of growth": divide everything by the tick dx that caused it. The change in area per unit of x is

\frac{d(uv)}{dx} = u\,\frac{dv}{dx} + v\,\frac{du}{dx} + \underbrace{\frac{du\,dv}{dx}}_{\to\; 0} = u\,v' + v\,u'.

The corner term dies in the limit — it's a product of two shrinking quantities divided by only one, so it vanishes as the step shrinks. What survives is exactly the product rule: one strip per factor, one term per strip. That's why there are two terms, and why each term pairs one changing factor (the derivative) with one factor held at its current size.

If you wrote u'v' on your first try, you're in the very best company. In November 1675, working through the brand-new calculus in his private notebooks, Gottfried Wilhelm Leibniz — one of its two inventors — jotted down the same guess: is d(uv) simply du \cdot dv? To his credit, he immediately asked whether it could actually be true, tested it on an example, and watched it fail — just as our x \cdot x check fails. Within days he had the correct d(uv) = u\,dv + v\,du, and it became one of the workhorse rules of his calculus. The wrong guess and its correction are both right there in the manuscripts: even the inventor had to be argued out of the "obvious" answer by a counterexample.

And what of the corner square we so breezily threw away? Making that rigorous is a lovely piece of limit work. Write the difference quotient for the product over a step h, then add and subtract the mixed term u(x+h)\,v(x) — the classic trick that splits one change into two:

\frac{u(x+h)v(x+h) - u(x)v(x)}{h} = u(x+h)\,\frac{v(x+h)-v(x)}{h} + v(x)\,\frac{u(x+h)-u(x)}{h}

Let h \to 0: the two fractions become v' and u' by definition, and u(x+h) \to u(x) because a differentiable function is continuous. The result is u\,v' + v\,u' — and notice the corner term never even shows up: the add-and-subtract split absorbed it. That "do no work, get it free" move is why the informal rectangle picture and the formal proof agree perfectly. Terms like the corner — proportional to h^2 when the strips are proportional to h — are exactly what limits are built to discard.

Worked example 1: a check against a rule you trust

Differentiate f(x) = (x^2)(x^3) with the product rule. Set out the four ingredients first — this small ritual prevents almost every product-rule error:

u = x^2,\quad u' = 2x, \qquad v = x^3,\quad v' = 3x^2

Assemble u'v + uv':

f'(x) = (2x)(x^3) + (x^2)(3x^2) = 2x^4 + 3x^4 = 5x^4

Now the reassuring cross-check: x^2 \cdot x^3 = x^5, and the power rule differentiates that directly to 5x^4. Two entirely different routes, the same answer. (Notice the wrong guess fails here too: u'v' = (2x)(3x^2) = 6x^3 — wrong power and wrong coefficient.)

Worked example 2: both roads, same destination

Something meatier: differentiate f(x) = (x^2 + 1)(x^3 - 2). This time each factor has more than one term — and we'll do it both ways, to build real trust in the rule.

Road A — the product rule. Ingredients:

u = x^2 + 1,\quad u' = 2x, \qquad v = x^3 - 2,\quad v' = 3x^2

Assemble and tidy:

f'(x) = (2x)(x^3 - 2) + (x^2 + 1)(3x^2) = 2x^4 - 4x + 3x^4 + 3x^2 = 5x^4 + 3x^2 - 4x

Road B — expand first, then differentiate term by term. Multiply out the product:

(x^2 + 1)(x^3 - 2) = x^5 + x^3 - 2x^2 - 2

Now the power rule on each term:

f'(x) = 5x^4 + 3x^2 - 4x - 0 = 5x^4 + 3x^2 - 4x \;\checkmark

Identical. For polynomials, the product rule and expand-first are always both available, and they always agree — the product rule is often less algebra (imagine expanding (x^4 + 3x^2 + 1)(x^5 - x^3 + 7) by hand versus writing down four ingredients). When you're unsure of a product-rule answer, expanding first is your built-in answer key.

Why the rule really matters: when expanding is impossible

If every product could be multiplied out, the product rule would just be a time-saver. Its real power shows when expanding is not an option at all. Consider

f(x) = x \sin x

There is no way to "multiply out" x and \sin x into a polynomial — they are different kinds of function, and the product is genuinely, irreducibly a product. Yet physics is full of exactly these: an oscillation whose amplitude grows with time is t \sin t; a signal that fades is a decaying function times a wave. The product rule handles them all with the same two-term recipe — once you know each factor's derivative. (You'll soon meet \frac{d}{dx}\sin x = \cos x, and then \frac{d}{dx}\big[x^2 \sin x\big] = 2x \sin x + x^2 \cos x falls out in one line.) That is the pattern for everything ahead: the product rule is the splitter that reduces a hard product to two easier derivatives, whatever those factors turn out to be.

Two traps account for nearly every lost mark on this rule:

u'v' is the classic wrong answer. It looks so plausible that it needs re-killing every few weeks. Keep the one-line antidote handy: for x \cdot x it predicts 1 \cdot 1 = 1, but the true derivative of x^2 is 2x. Whenever you're unsure whether you've remembered the rule correctly, run this thirty-second check before trusting your answer.
The rule is symmetric — but each term must pair ONE prime with ONE original. Writing u'v + uv' or uv' + u'v makes no difference (addition doesn't care about order, and neither does it matter which factor you call u). What does matter is the shape of each term: exactly one factor differentiated, the other left alone. If a term of your answer has two primes in it (u'v') or none (uv), it's wrong — no matter how the terms are ordered.

One more small comfort: you don't need the product rule for a constant multiple. \frac{d}{dx}\big[5x^2\big] is just 10x by the power rule — though if you insist on the product rule with u = 5, u' = 0 kills the first term and gives the same answer. The rule is never wrong to use; it's occasionally more machinery than the job needs.

Watch Sal explain it