Difference Quotients Before Limits, foundations for calculus lesson

Calculus defines the derivative as a limit of a fraction, and the limit step is usually short. The real work, and most of the mistakes, come earlier, in writing that fraction down and simplifying it. This page does only the finite part, with a real nonzero step $h$ , so that when limits arrive the algebra is already solid.

Key terms

Input change

Definition When the input of a function moves from $x_1$ to $x_2$ , the input change is

\Delta x = x_2 - x_1.

When the inputs are written as $a$ and $a + h$ , the input change is $(a + h) - a = h$ . It is positive when the input increases and negative when it decreases.

In plain words The input change is how far you step along the horizontal axis, with a sign that tells the direction.

Example Moving from $x = 5$ to $x = 3.5$ gives $\Delta x = 3.5 - 5 = -1.5$ . Starting at $a = 5$ , this is the step $h = -1.5$ .

Output change

Definition For a function $f$ and inputs $x_1$ and $x_2$ in its domain, the output change is

\Delta y = f(x_2) - f(x_1).

The order matters. Use the same order in the output change and the input change, the second input minus the first in both.

In plain words The output change is how far the graph rises or falls between the two inputs.

Example For $f(x) = x^2 + 3x$ , moving from $x = 1$ to $x = 2$ changes the output from $f(1) = 4$ to $f(2) = 10$ , so $\Delta y = 6$ .

Evaluating f(a + h) by substitution

Definition The expression $f(a + h)$ is the output of $f$ at the single input $a + h$ . It is found by substitution, replacing every occurrence of the input variable in the formula for $f$ by $(a + h)$ , with parentheses. In general

\begin{gathered} f(a + h) \ne f(a) + h \quad \text{and} \\[0.4em] f(a + h) \ne f(a) + f(h). \end{gathered}

In plain words Treat $a + h$ as one number and feed the whole thing into the formula. Parentheses keep the substitution honest.

Example For $f(x) = x^2$ and $a = 3$ ,

\begin{aligned} f(3 + h) &= (3 + h)^2 \\[0.4em] &= 9 + 6h + h^2. \end{aligned}

The tempting wrong answers are $f(3) + h = 9 + h$ and $f(3) + f(h) = 9 + h^2$ . Both miss the middle term $6h$ . With $h = 1$ the correct value is $f(4) = 16$ , while the wrong ones give $10$ .

Difference quotient

Definition Let $f$ be a real function, let $a$ and $h$ be real numbers with $h \ne 0$ , and suppose both $a$ and $a + h$ lie in the domain of $f$ . The difference quotient of $f$ at $a$ with step $h$ is

\begin{aligned} &\frac{f(a + h) - f(a)}{(a + h) - a} \\[0.6em] &\quad = \frac{f(a + h) - f(a)}{h}. \end{aligned}

The numerator is the output change and the denominator is the input change, taken in the same order.

In plain words It is the change in output divided by the change in input. It measures the average rate at which $f$ changes between $a$ and $a + h$ .

Example For $f(x) = x^2 + 3x$ and $a = 1$ we have $f(1) = 4$ and, by substitution,

\begin{aligned} f(1 + h) &= (1 + h)^2 + 3(1 + h) \\[0.4em] &= 4 + 5h + h^2. \end{aligned}

So the difference quotient is

\begin{aligned} \frac{f(1 + h) - f(1)}{h} &= \frac{5h + h^2}{h}, \\[0.4em] &\qquad h \ne 0. \end{aligned}

Secant slope

Definition The secant line through the distinct points $(a, f(a))$ and $(a + h, f(a + h))$ of the graph of $f$ has slope equal to the difference quotient

m_{\text{sec}} = \frac{f(a + h) - f(a)}{h}.

The step $h$ may be positive or negative. Listing the two points in the other order changes the sign of both the numerator and the denominator, so the slope does not depend on which point is called first.

In plain words The difference quotient is the slope of the straight line joining two points of the graph, rise over run.

Example For $f(x) = x^2 + 3x$ and $a = 1$ , take $h = 1$ and $h = -1$ .

$h$	Second point	Secant slope
$1$	$(2, 10)$	$\frac{10 - 4}{1} = 6$
$-1$	$(0, 0)$	$\frac{0 - 4}{-1} = 4$

These two plotted slopes will be used to check the simplified formula in the next section.

Cancellation

Definition A factor $h$ may be cancelled from a quotient only after it has been factored out of the entire numerator, and only for $h \ne 0$ . If the numerator equals $h \cdot g(h)$ , then

\frac{h \cdot g(h)}{h} = g(h) \quad \text{for } h \ne 0.

The simplified expression equals the difference quotient only at the values of $h$ where the difference quotient was defined. Cancelling a term that is added, rather than a common factor, is not allowed.

In plain words Pull an $h$ out of every term on top first. Then it can cancel with the $h$ underneath, and the answer comes with the reminder that $h$ is still not zero.

Example Continuing with $f(x) = x^2 + 3x$ at $a = 1$ ,

\begin{aligned} \frac{5h + h^2}{h} &= \frac{h(5 + h)}{h} \\[0.6em] &= 5 + h, \qquad h \ne 0. \end{aligned}

Check against the table. With $h = 1$ the formula gives $6$ and with $h = -1$ it gives $4$ , matching both secant slopes. A wrong move such as striking out the $h$ in $5h$ alone to get $\frac{5 + h^2}{h}$ would give $6$ at $h = 1$ but $-6$ at $h = -1$ , and the second check exposes it.

A quotient with fractions needs a common denominator first. For $g(x) = \frac{1}{x}$ at $a = 2$ , with $h \ne 0$ and $h \ne -2$ ,

\begin{aligned} g(2 + h) - g(2) &= \frac{1}{2 + h} - \frac{1}{2} \\[0.6em] &= \frac{2 - (2 + h)}{2(2 + h)} \\[0.6em] &= \frac{-h}{2(2 + h)}. \end{aligned}

Dividing by $h$ cancels the factor $h$ and gives

\frac{g(2 + h) - g(2)}{h} = -\frac{1}{2(2 + h)}.

The point $(4, \tfrac{1}{4})$ with $h = 2$ gives the secant slope $\frac{1/4 - 1/2}{2} = -\frac{1}{8}$ , and the formula gives $-\frac{1}{2 \cdot 4} = -\frac{1}{8}$ . The point $(1, 1)$ with $h = -1$ gives slope $\frac{1 - 1/2}{-1} = -\frac{1}{2}$ , and the formula gives $-\frac{1}{2 \cdot 1} = -\frac{1}{2}$ .

Nonzero increment and allowed steps

Definition The increment $h$ in a difference quotient must satisfy $h \ne 0$ , because the quotient divides by $h$ and the two points must be distinct. It may be positive or negative. It must also keep $a + h$ in the domain of $f$ , so the set of allowed $h$ depends on both the function and the point $a$ , and that restriction stays attached to every simplified form.

In plain words The step cannot be zero and cannot leave the domain. Every rewritten form of the quotient inherits those same rules.

Example For $g(x) = \frac{1}{x}$ at $a = 2$ , the input $2 + h$ must not be $0$ , so $h \ne -2$ , and $h \ne 0$ as always. The simplified form $-\frac{1}{2(2 + h)}$ happens to make sense at $h = 0$ , giving $-\frac{1}{4}$ , but the difference quotient itself does not exist there. That gap is exactly what the limit in calculus will deal with.

For $f(x) = x^2$ with general $a$ , the quotient is

\begin{aligned} \frac{(a + h)^2 - a^2}{h} &= \frac{2ah + h^2}{h} \\[0.6em] &= 2a + h, \\[0.4em] &\qquad h \ne 0. \end{aligned}

Every real $a$ and every nonzero real $h$ is allowed here, since $f$ is defined everywhere.

Check yourself

Find and simplify the difference quotient of $f(x) = 2x^2 - x$ at $a = 3$ . Check your answer with $h = 1$ .

Show the answer

Substitution gives

\begin{aligned} f(3 + h) &= 2(9 + 6h + h^2) \\[0.2em] &\quad - (3 + h) \\[0.4em] &= 15 + 11h + 2h^2, \end{aligned}

and $f(3) = 15$ . So

\begin{aligned} \frac{f(3 + h) - f(3)}{h} &= \frac{11h + 2h^2}{h} \\[0.6em] &= 11 + 2h, \\[0.4em] &\qquad h \ne 0. \end{aligned}

With $h = 1$ the formula gives $13$ . The points $(3, 15)$ and $(4, 28)$ give slope $13$ as well.

Simplify the difference quotient of $f(x) = \frac{1}{x + 1}$ at $a = 1$ and state which values of $h$ are allowed.

Show the answer

The common denominator gives

\frac{1}{2 + h} - \frac{1}{2} = \frac{-h}{2(2 + h)},

since $f(1 + h) = \frac{1}{2 + h}$ and $f(1) = \frac{1}{2}$ . Dividing by $h$ cancels the factor $h$ , so the difference quotient is $-\frac{1}{2(2 + h)}$ . The input $1 + h$ must not be $-1$ , so $h \ne -2$ , and also $h \ne 0$ .

For $f(x) = \sqrt{x}$ at $a = 4$ , which steps $h$ are allowed in the difference quotient?

Show the answer

The input $4 + h$ must be in the domain $[0, \infty)$ , so $h \ge -4$ . Together with $h \ne 0$ , the allowed steps are $-4 \le h < 0$ and $h > 0$ .

Difference Quotients Before Limits

Key terms

Input change

Output change

Evaluating f(a + h) by substitution

Difference quotient

Secant slope

Cancellation

Nonzero increment and allowed steps

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus

Resources for this lesson

Tangent line explorer