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Pure · Differentiation

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The product rule

Differentiating a product of two functions, y = uv, as u'v + uv' — why the naive "differentiate each and multiply" is wrong, and how to combine the rule with the chain rule.

Pure · Differentiation

The product rule

Differentiating a product of two functions, y = uv, as u'v + uv' — why the naive "differentiate each and multiply" is wrong, and how to combine the rule with the chain rule.

Why it works

When two functions are multiplied — say y=x2(2x+1)3y = x^2(2x+1)^3 — the temptation is to differentiate each factor and multiply the results. That is wrong, and it is worth seeing why before learning the fix.

A product measures an area. Picture a rectangle whose width is uu and height is vv, so its area is y=uvy = uv. Nudge xx a little: the width grows by a sliver δu\delta u and the height by a sliver δv\delta v. The area grows by three pieces — a tall thin strip v δuv\,\delta u down the side, a long thin strip u δvu\,\delta v along the top, and a tiny corner rectangle δu δv\delta u\,\delta v. That corner is the product of two small things, so it is vanishingly small compared with the strips and disappears in the limit. The change in area is therefore δy≈v δu+u δv.\delta y \approx v\,\delta u + u\,\delta v. Divide by δx\delta x and let it shrink to zero: dydx=udvdx+vdudx.\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}. The reason "multiply the two derivatives" fails is now clear — it keeps only the corner and throws away the two strips that actually matter.

The rule, in the form to memorise: if y=uvy = uv then dydx=u′v+uv′.\frac{dy}{dx} = u'v + uv'. Differentiate the first times the second, plus the first times the derivative of the second. Set the work out by naming uu and vv, writing down u′u' and v′v' underneath, and assembling. Order does not matter — addition is commutative — but a tidy layout prevents slips.

Very often one factor is itself a composite, so you need the chain rule inside the product rule. For y=x2(2x+1)3y = x^2(2x+1)^3, take u=x2u = x^2 and v=(2x+1)3v = (2x+1)^3; then u′=2xu' = 2x and v′=3(2x+1)2×2=6(2x+1)2v' = 3(2x+1)^2\times 2 = 6(2x+1)^2 (chain rule). Assemble and, where the examiner expects it, factorise the answer — a fully factorised derivative is what you need to solve dydx=0\frac{dy}{dx} = 0 later.