Product Rule

Multiplying Functions Before Differentiating

The Sum Rule separates the change in a sum into the changes in its two terms. For a product, the change is more involved: when both factors vary, the change in their product reflects changes in each factor. The Product Rule gives a precise formula for this effect. Its hypotheses, like those of the Sum Rule, are local: the functions need only be differentiable at the point where the derivative is being computed.

Let $I$ be an interval, and let $f,g:I\to\mathbb{R}$. Their product is the function $fg$ defined by $(fg)(x)=f(x)g(x)$. For an interior point $a\in I$, the rule expresses the derivative of $fg$ at $a$ using the values and derivatives of $f$ and $g$ at $a$.

Theorem (Product Rule): Let $I$ be an interval, let $a$ be an interior point of $I$, and let $f,g:I\to\mathbb{R}$ be differentiable at $a$. Then $fg$ is differentiable at $a$, and $$ (fg)'(a)=f'(a)g(a)+f(a)g'(a). $$

The formula has two terms. The first records the change in $f$, weighted by the value of $g$; the second records the change in $g$, weighted by the value of $f$. In general, both terms are necessary. A useful exact identity for the difference quotient is

\frac{f(a+h)g(a+h)-f(a)g(a)}{h} = \frac{f(a+h)-f(a)}{h}g(a+h) + f(a)\frac{g(a+h)-g(a)}{h}.

To verify it, expand the right-hand side: its numerator after combining terms is $f(a+h)g(a+h)-f(a)g(a+h)+f(a)g(a+h)-f(a)g(a)$. The two middle terms cancel, leaving exactly the numerator on the left. As $h\to0$, the first difference quotient tends to $f'(a)$, the second to $g'(a)$, and $g(a+h)$ tends to $g(a)$ because differentiability implies continuity. This identity shows how the two terms in the rule arise. The first quotient is multiplied by a value of $g$ that varies with $h$, so controlling that factor is part of a complete limit argument.

A Useful Local Bound

The Product Rule’s difference-quotient reasoning uses the fact that a differentiable function cannot become arbitrarily large arbitrarily close to the point of differentiation. We establish this local bound directly. It is a useful technique whenever a difference quotient contains a factor whose limit is known to be finite.

Theorem (Local Boundedness from Differentiability): If $f:I\to\mathbb{R}$ is differentiable at an interior point $a$ of an interval $I$, then there are numbers $\delta>0$ and $M\geq0$ such that $|f(x)|\leq M$ whenever $x\in I$ and $|x-a|<\delta$.

Proof. Differentiability means that the difference quotient $(f(a+h)-f(a))/h$ tends to the finite number $f'(a)$ as $h\to0$. By the definition of a finite limit, there is a $\delta_1>0$ such that, for $0<|h|<\delta_1$ with $a+h\in I$,

\left|\frac{f(a+h)-f(a)}{h}-f'(a)\right|<1.

It follows from the triangle inequality that

\left|\frac{f(a+h)-f(a)}{h}\right|<|f'(a)|+1.

Therefore, for these $h$,

|f(a+h)| \leq |f(a)|+|f(a+h)-f(a)| \leq |f(a)|+|h|(|f'(a)|+1).

Choose $\delta>0$ with $\delta\leq\delta_1$ and $\delta\leq1$. Then for $0<|h|<\delta$, the last expression is at most $|f(a)|+|f'(a)|+1$. At $h=0$, $|f(a)|$ satisfies the same bound. Thus $M=|f(a)|+|f'(a)|+1$ works for all $x=a+h\in I$ with $|h|<\delta$. This proves local boundedness. $\square$

In the Product Rule identity, differentiability of $g$ gives continuity at $a$, hence $g(a+h)\to g(a)$. Local boundedness is a related safeguard: it ensures that values near $a$ remain finite and controlled. The two ideas often accompany one another in limit arguments, but boundedness alone does not supply the limit of $g(a+h)$.

When One Factor Vanishes at the Point

If one factor equals zero at $a$, the general product formula simplifies. This special case can also be proved directly from the definition, and it illustrates how a value at the point affects the derivative.

Theorem (Product Rule at a Zero): Let $f,g:I\to\mathbb{R}$ be differentiable at an interior point $a\in I$. If $f(a)=0$, then $fg$ is differentiable at $a$ and $$ (fg)'(a)=f'(a)g(a). $$

Proof. Since $f(a)=0$, we have $(fg)(a)=0$. For nonzero $h$ small enough that $a+h\in I$, the difference quotient is

\frac{(fg)(a+h)-(fg)(a)}{h} = \frac{f(a+h)g(a+h)}{h} = \frac{f(a+h)-f(a)}{h}g(a+h).

The first factor tends to $f'(a)$ by differentiability of $f$. Differentiability of $g$ implies continuity at $a$, so the second factor tends to $g(a)$. The product law for limits shows that the displayed difference quotient tends to $f'(a)g(a)$. This finite limit is the derivative of $fg$ at $a$, proving the claim. $\square$

This special case does not require $g(a)=0$; it is enough that $f(a)=0$. If instead $g(a)=0$, the same argument with the roles reversed gives $(fg)'(a)=f(a)g'(a)$. When both factors vanish, both expressions give zero.

Worked Example: A Product of Two Polynomials

Let $f(x)=x^3-2x$ and $g(x)=x^2+1$. The Power Rule and the Sum Rule give $f'(x)=3x^2-2$ and $g'(x)=2x$. At $a=1$,

f(1)=1-2=-1,\qquad g(1)=1+1=2,

and

f'(1)=3-2=1,\qquad g'(1)=2.

The Product Rule therefore gives

$$ (fg)'(1)=f'(1)g(1)+f(1)g'(1) =(1)(2)+(-1)(2)=0. $$

As a check, multiplying first gives $f(x)g(x)=x^5-x^3-2x$. Its polynomial derivative is $5x^4-3x^2-2$, which at $1$ equals $5-3-2=0$, in agreement with the rule.

Worked Example: A Polynomial Times a Reciprocal

On $I=(0,\infty)$, define $f(x)=x^2+3$ and $g(x)=1/x$. Both are differentiable at every point of $I$, with $f'(x)=2x$ and $g'(x)=-1/x^2$. At $a=2$, the values are

f(2)=7,\qquad g(2)=\frac12,\qquad f'(2)=4,\qquad g'(2)=-\frac14.

Consequently,

(fg)'(2) =4\left(\frac12\right)+7\left(-\frac14\right) =2-\frac74 =\frac14.

The domain restriction matters: $1/x$ is not defined at zero. Here the Product Rule is applied at $2$, an interior point of the interval on which both functions are defined and differentiable.

Worked Example: Using the Zero-Value Case

Define $f(x)=x^2-4$ and $g(x)=x^3+1$. At $a=2$, $f(2)=4-4=0$, while $g(2)=8+1=9$. The derivatives are $f'(x)=2x$ and $g'(x)=3x^2$, so $f'(2)=4$. The Product Rule at a Zero gives

(fg)'(2)=f'(2)g(2)=4\cdot9=36.

Using the full formula gives the same result: $f'(2)g(2)+f(2)g'(2)=4\cdot9+0\cdot12=36$. The zero factor removes one term, but it does not make the derivative of the product zero unless the other term also vanishes.

Using the Rule Carefully

The Product Rule is not the statement $(fg)'=f'g'$. The derivatives must be paired with the opposite function values: one term is $f'g$, and the other is $fg'$. The rule also requires both functions to be differentiable at the point. Continuity of the factors by itself does not guarantee that their product is differentiable.

Conversely, differentiability of a product does not imply that each factor is differentiable. For instance, $u(x)=|x|$ is not differentiable at zero, as its difference quotient has value $1$ for positive $h$ and $-1$ for negative $h$. Yet $u(x)u(x)=x^2$, which is differentiable at zero by the Power Rule. This does not contradict the Product Rule: the rule guarantees differentiability of a product when both factors are differentiable; it does not claim that this hypothesis is necessary in every case.

For a calculation, a reliable procedure is to identify the two factors, compute each factor’s value and derivative at the point, and then form both terms in the formula. Keeping the terms separate until the final arithmetic makes it less likely that a factor value or derivative will be omitted. The difference-quotient identity explains why this two-term structure is necessary, while the zero-value case offers a quick simplification when one factor vanishes.

Check Your Understanding

Use the Product Rule, the local boundedness result, and the difference quotient where appropriate.

State the hypotheses of the Product Rule and write its formula at $a$.
If $f'(a)=3$, $g(a)=-2$, $f(a)=4$, and $g'(a)=5$, find $(fg)'(a)$.
If $f(a)=0$, $f'(a)=-2$, and $g(a)=7$, find $(fg)'(a)$ using the zero-value case.
Why does differentiability of $f$ at $a$ imply that $f$ is bounded in some neighborhood of $a$?
Give an example of a product that is differentiable at zero even though one of its factors is not, and explain why it does not contradict the Product Rule.

What You'll Learn