Product and Quotient Rules

Why products need two contributions

When both factors change, a product changes through either factor. Writing the increment of $u(x)v(x)$ and adding/subtracting $u(x+h)v(x)$ splits the difference quotient into two terms. Taking limits gives

(uv)'=u'v+uv'.

Each term changes one factor while retaining the other. Multiplying derivatives would discard most of the change.

For a quotient with $v\ne0$ ,

\left(\frac uv\right)'=\frac{u'v-uv'}{v^2}.

The numerator's order matters. One way to derive the rule is to differentiate $u=v(u/v)$ using the product rule and solve for the unknown derivative.

Visual guide

VISUAL GUIDEChanging a rectangle changes two strips

When both sides x and y change, first-order area change is y·dx + x·dy. The small corner dx·dy is second order. This is the geometric reason for the product rule’s two terms.

Worked example: a product with competing rates

For $f(x)=x^2\sin x$ , take $u=x^2$ and $v=\sin x$ . Then

f'(x)=2x\sin x+x^2\cos x.

At $x=\pi$ , the derivative is $-\pi^2$ . Although $x^2$ is increasing for positive $x$ , the product can decrease because the sine factor changes. A derivative of a product is not determined by one factor alone.

If the function were $x^2(x+1)$ , expanding to $x^3+x^2$ would be simpler. The product rule gives $2x(x+1)+x^2$ , which agrees with $3x^2+2x$ . Use algebra to choose the shortest transparent method.

Worked example: a rational function

For $g(x)=(x^2+1)/(x-1)$ , the domain excludes $1$ . The quotient rule gives

g'(x)=\frac{2x(x-1)-(x^2+1)}{(x-1)^2}=\frac{x^2-2x-1}{(x-1)^2}.

As a check, polynomial division gives $g=x+1+2/(x-1)$ . Differentiating yields $1-2/(x-1)^2$ , the same result. Two independent forms can catch a sign mistake.

A useful special case

For $1/v(x)$ , the numerator derivative vanishes, leaving $-v'/v^2$ . This explains the negative sign for reciprocal differentiation. If a simplification cancels a factor, retain any exclusions inherited from the original expression; an algebraically removable hole still belongs to the original problem.

A derivative can itself have zeros or undefined points, but those need separate interpretation. Solving the numerator of $g'$ for zero is meaningful only at inputs in the original domain.

Practice

Differentiate $(x+2)e^x$ .
Differentiate $x/(x+1)$ .
Differentiate $x^2/x$ on its original domain.

Show worked solutions

The product rule gives $e^x+(x+2)e^x=(x+3)e^x$ .
The numerator is $(x+1)-x=1$ , giving $1/(x+1)^2$ for $x\ne-1$ .
The function equals $x$ only for $x\ne0$ , so its derivative is $1$ there. The original function still has no derivative at zero because it is not defined there.