Sum, Product, and Quotient Rules

First principles establish what a derivative means. They are less convenient for differentiating a complicated expression repeatedly. We can prove rules once, then use them whenever their hypotheses hold. Every rule in this chapter assumes the component functions are differentiable at the point in question. The quotient rule also requires a nonzero denominator.

Linearity comes from the limit laws

For constants $c,d$, expand the difference quotient of $cf+dg$ and split it into two quotients. Taking limits gives

$$(cf+dg)'=cf'+dg'.$$

Thus $(4x^3-2x^2+7)'=12x^2-4x$. The constant contributes zero. Differentiating the square of a function is not covered by linearity; squaring is multiplication, not addition.

Why both factors contribute

For $F=fg$, write the numerator as

$$f(x+h)g(x+h)-f(x)g(x) =f(x+h)[g(x+h)-g(x)]+g(x)[f(x+h)-f(x)].$$

After division by $h$, take limits. Differentiability implies $f(x+h)\to f(x)$, so

$$(fg)'=fg'+gf'.$$

The extra term that a naive “differentiate both factors and multiply” approach loses records the change in one factor while the other remains present. For $x\cdot x$, the rule gives $x+x=2x$, whereas multiplying the two derivatives gives $1$, which fails even this basic check.

For $F(x)=(x^2+1)(x^3-2)$,

$$F'=2x(x^3-2)+(x^2+1)3x^2=5x^4+3x^2-4x.$$

Expanding first gives $F=x^5+x^3-2x^2-2$ and the same derivative. Independent representations are useful error checks. They are often more informative than repeating the same calculation with slightly different notation.

The quotient rule

For a differentiable nonzero $g$, the difference quotient for $1/g$ simplifies to

$$-\frac{g(x+h)-g(x)}{h\,g(x+h)g(x)}.$$

Its limit is $-g'/g^2$. Continuity and $g(x)\ne0$ ensure $g(x+h)$ stays nonzero for sufficiently small increments. Applying the product rule to $f(1/g)$ gives

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

The order in the numerator matters. For $Q(x)=(x^2+1)/(x-1)$,

$$Q'(x)=\frac{2x(x-1)-(x^2+1)}{(x-1)^2} =\frac{x^2-2x-1}{(x-1)^2},\quad x\ne1.$$

Polynomial division gives $Q=x+1+2/(x-1)$, whose derivative is $1-2/(x-1)^2$, algebraically identical to the result. The forbidden input $1$ remains forbidden in both versions.

Negative integer powers

For a positive integer $n$, apply the reciprocal rule to $x^n$:

$$\frac d{dx}x^{-n}=-\frac{nx^{n-1}}{x^{2n}}=-nx^{-n-1},\qquad x\ne0.$$

The power rule therefore holds for all integer exponents on their relevant domains. At exponent zero the function is constant wherever it is defined. This extension came from a new argument, not from silently expanding the earlier binomial proof beyond its assumptions.

Choosing a method

First identify the outermost operation. An expression can contain sums inside a product inside a quotient. Start with the outside, retain its component functions intact, then differentiate those components. Simplify when simplification reduces error; leaving a correctly factored derivative is often preferable to an unnecessarily long polynomial.

Numerical difference quotients at ordinary inputs can detect a sign error. Very small increments may instead amplify floating-point cancellation. A numerical agreement supports a check at that input; the proved rules establish the symbolic result across the domain.

Exercises

1. Differentiate $(2x-1)(x^2+3)$ by the product rule and by expansion.
2. Differentiate $(x+1)/(x+2)$ and state its domain.
3. Differentiate $x^{-3}+5x^2-8$.
4. Explain why $(f/g)'=f'/g'$ cannot be a general rule.
5. Derive the rule for a product of three differentiable functions using the two-factor rule twice.
6. Find the tangent slope of $(x^2+1)/(x-1)$ at $x=0$.

Worked solutions

1. $2(x^2+3)+(2x-1)2x=6x^2-2x+6$. Expanding first gives $2x^3-x^2+6x-3$ and the same derivative.
2. $[(x+2)-(x+1)]/(x+2)^2=1/(x+2)^2$, for $x\ne-2$.
3. $-3x^{-4}+10x$, for $x\ne0$.
4. Take $f=g=x$ away from zero. The quotient is constant with derivative zero, while $f'/g'=1$.
5. $(fgh)'=f'gh+fg'h+fgh'$.
6. Substitute into the verified derivative: $(-1)/1=-1$.