The Product Rule

A product as a rectangle

Let y = uv, where u and v are both functions of x. Picture the product as the area of a rectangle, u wide and v tall. When x grows by a small step $\Delta x$Δx, u grows by $\Delta u$Δu and v grows by $\Delta v$Δv, and the rectangle grows to $\left(u+\Delta u\right)\left(v+\Delta v\right)$(u + Δu)(v + Δv).

The new area is the old one plus three pieces: a strip $v\cdot \Delta u$v·Δu down one side, a strip $u\cdot \Delta v$u·Δv along the top, and a small corner $\Delta u\cdot \Delta v$Δu·Δv where the two strips meet.

A u by v rectangle grown a little in each direction. It gains a strip along each side, $u\cdot \Delta v$u·Δv and $v\cdot \Delta u$v·Δu, and a small corner $\Delta u\cdot \Delta v$Δu·Δv.

The corner vanishes

Multiplying out, $\left(u+\Delta u\right)\left(v+\Delta v\right)=uv+v\cdot \Delta u+u\cdot \Delta v+\Delta u\cdot \Delta v$(u + Δu)(v + Δv) = uv + v·Δu + u·Δv + Δu·Δv, so the growth in y is $\Delta y=v\cdot \Delta u+u\cdot \Delta v+\Delta u\cdot \Delta v$Δy = v·Δu + u·Δv + Δu·Δv. Divide by the step $\Delta x$Δx: $\Delta y/\Delta x=v\times \left(\Delta u/\Delta x\right)+u\times \left(\Delta v/\Delta x\right)+\left(\Delta u/\Delta x\right)\times \Delta v$Δy / Δx = v × (Δu / Δx) + u × (Δv / Δx) + (Δu / Δx) × Δv.

Now let $\Delta x$Δx tend to 0. $\Delta u/\Delta x$Δu / Δx tends to u' and $\Delta v/\Delta x$Δv / Δx tends to v'. In the last term, $\Delta u/\Delta x$Δu / Δx tends to u' but $\Delta v$Δv tends to 0, because v has a derivative and is therefore continuous. So the corner term tends to u' × 0 = 0. What is left is the product rule: (uv)' = u'v + uv'.

In words: differentiate u and keep v, then keep u and differentiate v, and add. In Leibniz's notation, $\frac{dy}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}$dy/dx = v du/dx + u dv/dx.

A 4 by 3 rectangle grown by $\Delta u=1.6$Δu = 1.6 and $\Delta v=1.6$Δv = 1.6. The strips are $u\cdot \Delta v=6.4$u·Δv = 6.4 and $v\cdot \Delta u=4.8$v·Δu = 4.8, and the corner is 1.6 × 1.6 = 2.56, about 18.6% of the growth. Drag both handles down to 0.2: the strips shrink to 0.8 and 0.6, but the corner, a small amount times a small amount, shrinks to 0.04, under 3% of the growth.

A check against the power rule

Take $y=x\cdot x^3$y = x · x³, with u = x and $v=x^3$v = x³. Then u' = 1 and $v^{\prime }=3x^2$v' = 3x², so the product rule gives $1\cdot x^3+x\cdot 3x^2=x^3+3x^3=4x^3$1 · x³ + x · 3x² = x³ + 3x³ = 4x³. The product is $x^4$x⁴, and the power rule gives $4x^3$4x³ too.

Take $y=x^2\cdot x^3$y = x² · x³, with $u=x^2$u = x² and $v=x^3$v = x³. The rule gives $2x\cdot x^3+x^2\cdot 3x^2=2x^4+3x^4=5x^4$2x · x³ + x² · 3x² = 2x⁴ + 3x⁴ = 5x⁴. The product is $x^5$x⁵, and the power rule gives $5x^4$5x⁴ again.

Not the product of the derivatives

It is tempting to differentiate each factor and multiply: (uv)' = u'v'. That is wrong, and one example is enough to show it. For $x^2\cdot x^3$x² · x³, the derivatives of the factors are 2x and $3x^2$3x², whose product is $6x^3$6x³. At x = 2 that is 48. But $x^2\cdot x^3=x^5$x² · x³ = x⁵, whose derivative $5x^4$5x⁴ is 80 at x = 2, and the chord from x = 2 with h = 0.001 gives $\left(2.001^5-32\right)/0.001=80.08004$(2.001⁵ − 32) / 0.001 = 80.08004. The answer is 80, not 48.

The simplest case shows it too. For $x\cdot x=x^2$x · x = x², the derivative is 2x, but 1 × 1 = 1. Each term of the product rule keeps one factor whole, and the product of the two derivatives keeps neither.

The rule works for any two factors that have derivatives, such as $x^2$x² and sin x on the rectangle below: once the derivative of each factor is known, the derivative of the product is the first factor's derivative times the second factor, plus the first factor times the second factor's derivative.

A square of side x, so u = v = x and $uv=x^2$uv = x². Growing x by $\Delta x$Δx adds two strips, each $x\cdot \Delta x$x·Δx, so the true rate is 2x. The u'v' guess, 1 × 1 = 1, accounts for one unit of strip, shaded on its own. Drag x to the right: the true rate grows with x and the guess stays at 1.

The rectangle with sides $x^2$x² and sin x. When x grows, one strip is sin x times the growth of $x^2$x², and the other is $x^2$x² times the growth of sin x: two terms, one for each factor.

Products of polynomials and roots

For $y=\left(x^2+1\right)\left(3x-2\right)$y = (x² + 1)(3x − 2), take $u=x^2+1$u = x² + 1 and v = 3x − 2, so u' = 2x and v' = 3. The rule gives $2x\left(3x-2\right)+3\left(x^2+1\right)=6x^2-4x+3x^2+3=9x^2-4x+3$2x(3x − 2) + 3(x² + 1) = 6x² − 4x + 3x² + 3 = 9x² − 4x + 3. Check by expanding first: $y=3x^3-2x^2+3x-2$y = 3x³ − 2x² + 3x − 2, which differentiates to $9x^2-4x+3$9x² − 4x + 3. At x = 1 that is 8, and the chord with h = 0.001 gives 8.007003.

For $y=\sqrt{x}\left(x+1\right)$y = √x (x + 1), take $u=\sqrt{x}$u = √x and v = x + 1, so $u^{\prime }=\frac{1}{2\sqrt{x}}$u' = 1/(2√x) and v' = 1. The rule gives $\frac{x+1}{2\sqrt{x}}+\sqrt{x}$(x + 1)/(2√x) + √x. At x = 4 that is $\frac{5}{4}+2=3.25$5/4 + 2 = 3.25, and the chord with h = 0.001 gives 3.250172.

A constant factor is a factor too. For $5x^3$5x³, with u = 5 and $v=x^3$v = x³, the rule gives $0\cdot x^3+5\cdot 3x^2=15x^2$0 · x³ + 5 · 3x² = 15x², the same as keeping the 5 as a multiplier. And three factors give three terms, each with one factor differentiated: for $x\cdot x\cdot x=x^3$x · x · x = x³ that is $1\cdot x\cdot x+x\cdot 1\cdot x+x\cdot x\cdot 1=3x^2$1 · x · x + x · 1 · x + x · x · 1 = 3x².

The usual mistakes

Multiplying the two derivatives. For $x\cdot x^2$x · x² that gives 1 × 2x = 2x, but the derivative of $x^3$x³ is $3x^2$3x²: $1\cdot x^2+x\cdot 2x$1 · x² + x · 2x.

Writing only one term. For $x\cdot x^3$x · x³, differentiating $x^3$x³ alone gives $3x^2$3x²; the factor x contributes its own term, $1\cdot x^3$1 · x³, and the answer is $4x^3$4x³.

Subtracting the two terms. A minus sign belongs to the quotient rule; for a product, both pieces of growth add.

Leaving the product undifferentiated: answering $x^4$x⁴ for the derivative of $x\cdot x^3$x · x³. That is the product itself.

A bakery

In the application below, a bakery lowers its price and sells more loaves each week, so its takings are a product of two changing quantities, price times loaves sold. The product rule gives the rate the takings change, and setting it to 0 finds the week the takings are steady.