The Difference Quotient
An average-speed camera clocks you by timing how long you take to cross a short painted stretch
of road. Shorten that stretch — a hundred metres, then ten, then one — and the figure closes in
on your speed at a single instant. The difference quotient is exactly that measurement
written as one tidy formula, with a dial for the length of the stretch so we can turn it right
down.
The average rate of
change needs two input values, a and
b — two separate anchors, planted wherever you like. That's fine for
answering "how fast, on average, between here and there?" But the last page ended with a
different, itchier question: what happens as the two points slide together? To chase
that, we want a formula built for shrinking — one where the whole interval is
controlled by a single number we can squeeze.
So here is the one small, brilliant change of notation. Keep the first point and call it
x. Then, instead of naming the second point independently, describe it
by how far past the first it sits: step forward by h, landing
at x + h. Using
function notation, the
average rate of change between those two inputs becomes the difference quotient:
\frac{f(x + h) - f(x)}{(x + h) - x} = \frac{f(x + h) - f(x)}{h}
The run is simply h — the second input minus the first collapses to
the step size — and the numerator is the rise: how much the output changed over that step.
Nothing new has been computed. This is exactly the same fraction as
\tfrac{f(b) - f(a)}{b - a} with b = x + h.
But the new costume matters enormously: the width of the stretch is now a single dial,
h, and a dial can be turned down. One formula now
describes every secant through the point \big(x, f(x)\big) —
the wide ones, the narrow ones, and (soon) the vanishingly thin ones. The difference quotient is
the derivative's chrysalis: everything the derivative will be is already folded up inside it,
waiting for h to shrink.
Getting to know h
Three quick facts about the new dial, worth fixing in place before the algebra starts.
h is a width, not a location. It measures how far the
second point sits from the first. h = 2 is a wide stretch;
h = 0.01 is a sliver. And h may be
negative: h = -1 just means the second point is one unit to
the left, and the same fraction still computes the secant's slope, sign and all.
The one forbidden value is h = 0. Set
h = 0 and both points coincide; the fraction reads
\tfrac{f(x) - f(x)}{0} = \tfrac{0}{0}, which is meaningless. A secant
needs two distinct points. The entire drama of the next few pages is about sneaking up on
h = 0 without ever standing on it.
It is a machine with two dials. Feed it a base point
x and a step h; it returns the slope of
the secant from \big(x, f(x)\big) to
\big(x+h, f(x+h)\big). Different x,
different part of the curve; different h, different width of view.
It is the slope of a secant
Geometrically, the difference quotient is the slope of the secant line joining
\big(x, f(x)\big) to
\big(x + h, f(x + h)\big). Below, the base point is pinned at
x = 1 on f(x) = x^2, and the slider is the
h dial. Slide it and watch the second point travel along the curve
while the secant tilts to match.
Two experiments worth running. First, park h at
1 and note the secant's steepness — we'll compute below that its slope
is exactly 2x + h = 2(1) + 1 = 3. Then drag
h slowly down toward the left end. The two points crowd together, the
secant stops swinging, and its tilt settles toward one particular steepness — the slope readings
3, 2.5, 2.1, \dots are closing in on 2.
The slider physically cannot reach h = 0 (that's the forbidden value),
but the secant's destination is already unmistakable. Hold that thought.
Worked example: simplify it for f(x) = x^2
The real power of the difference quotient is that the algebra often simplifies — and the whole
point of the exercise is to simplify it completely, until the lone
h downstairs is gone. Watch it happen for
f(x) = x^2, honestly, with no steps skipped.
Step 1 — build f(x+h). Substitute
(x+h) for the input and expand the square in full:
f(x + h) = (x + h)^2 = (x+h)(x+h) = x^2 + 2xh + h^2.
Step 2 — subtract f(x). The
x^2 terms annihilate each other — this cancellation is the engine of
the whole method, and it happens for every function you'll meet:
f(x+h) - f(x) = (x^2 + 2xh + h^2) - x^2 = 2xh + h^2.
Step 3 — divide by h. Every surviving term in the
numerator carries a factor of h, so factor it out and cancel (legal
precisely because h \neq 0):
\frac{2xh + h^2}{h} = \frac{h(2x + h)}{h} = 2x + h.
Pause and admire what just happened. The pesky h in the denominator
is gone. The slope of any secant on the parabola — any base point, any width —
is the clean expression 2x + h. From
x = 1 with step h = 1: slope
2(1) + 1 = 3. From x = 3 with step
h = 0.5: slope 6.5. Infinitely many
secants, one formula. And crucially, the formula no longer objects to small
h: where \tfrac{2xh + h^2}{h} degenerates
into \tfrac{0}{0} at h = 0, the simplified
2x + h is perfectly happy there. That is why the simplification is
the whole game.
See the algebra animated
The numerator and denominator change together as the step h
plays out. Press play to build the difference quotient for
f(x) = x^2 at x = 1: the step along the
axis is the run h, the vertical leg is the rise
f(x+h) - f(x), and their ratio is the secant's slope —
2x + h, exactly as the algebra promised.
Worked example: a straight line, f(x) = 3x + 1
Now run the same three steps on a function whose answer you already know. A line has one slope,
everywhere — so its average rate of change over any stretch had better come out the
same. Does the machinery agree?
Step 1: substitute and expand (the input (x+h) replaces every x):
f(x+h) = 3(x+h) + 1 = 3x + 3h + 1.
Step 2: subtract — and this time almost everything cancels:
f(x+h) - f(x) = (3x + 3h + 1) - (3x + 1) = 3h.
Step 3: divide:
\frac{3h}{h} = 3.
Just 3. No x, no
h — the answer doesn't care where you stand or how wide you look,
which is precisely what "a line has constant slope" means. (The +1
never stood a chance: any added constant appears in both f(x+h) and
f(x) and is wiped out by the subtraction. Shifting a graph up or down
changes none of its slopes.) The extreme case is a constant function
f(x) = k: numerator k - k = 0, difference
quotient 0. A flat graph has zero rate of change over every stretch —
the machine confirms the obvious, which is exactly what a machine should do before you trust it
on the non-obvious.
Turn the dial: watch 2x + h as h shrinks
Back to the parabola, armed with the simplified formula. Fix the base point at
x = 1, so every secant slope is 2(1) + h = 2 + h,
and now do the thing the notation was built for — turn the h
dial down and just read off the answers:
| step h |
secant slope 2 + h |
| 1 |
3 |
| 0.5 |
2.5 |
| 0.1 |
2.1 |
| 0.01 |
2.01 |
| 0.001 |
2.001 |
No fresh computation, no new function values, no wrestling with tiny numerators over tiny
denominators — the unsimplified fraction would have had you dividing
0.002001 by 0.001 at the last row. Instead
the pattern is naked: 3,\ 2.5,\ 2.1,\ 2.01,\ 2.001,\ \dots The slopes
are converging on 2, and the formula says why: as
h melts away, 2 + h can only go to
2. What you are watching emerge from that table is the
derivative of x^2 at x = 1
— the parabola's exact steepness at a single point. We haven't formally defined it yet (that's
the limit's job,
coming shortly), but the difference quotient has already handed us the number.
Worked example: a fraction, f(x) = \dfrac{1}{x}
One more, because not every function is a polynomial and the cancellation doesn't always arrive
by expanding brackets. For f(x) = \tfrac{1}{x} the tool you need is an
old friend: common denominators.
Step 1: substituting is easy here — the whole input drops into the denominator:
f(x+h) = \frac{1}{x+h}.
Step 2: subtract. Two fractions with different denominators want a common one,
namely x(x+h):
\frac{1}{x+h} - \frac{1}{x} = \frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}.
Watch the top line closely: x - (x+h). The
x's cancel and only -h survives — there's
the factor of h again, delivered by subtraction rather than expansion.
(Mind the minus sign: subtracting the whole bracket gives
x - x - h, not x - x + h.)
Step 3: divide by h — dividing a fraction by h multiplies its denominator:
\frac{f(x+h)-f(x)}{h} = \frac{-h}{x(x+h)} \cdot \frac{1}{h} = \frac{-1}{x(x+h)}.
Fully simplified: \dfrac{-1}{x(x+h)}. The bare
h in the denominator is gone (the x + h
that remains is harmless — it doesn't vanish when h does), and the
formula already whispers the ending: shrink h and the secant slopes
head for \tfrac{-1}{x \cdot x} = -\tfrac{1}{x^2}. Negative, as it
should be — the graph of \tfrac{1}{x} falls as
x grows. Three functions, three different kinds of algebra
(expanding, cancelling, common denominators), one ritual: build, subtract, divide, and
simplify until the lone h downstairs is gone.
Difference-quotient homework is graded around the world every day, and the same two mistakes
account for most of the red ink.
-
f(x+h) means substitute (x+h) for
every x — then expand properly. The single most
common error in all of introductory calculus is writing
(x+h)^2 = x^2 + h^2. It is not:
(x+h)^2 = (x+h)(x+h) = x^2 + 2xh + h^2, and that middle term
2xh is not a technicality — it is the term that survives
the subtraction and becomes the derivative. Drop it and the whole computation collapses to
nonsense. Related impostors to refuse on sight: f(x+h) is not
f(x) + h, and not f(x) + f(h). When in
doubt, write the function with an empty box — f(\square) = \square^2
— and drop the entire parcel (x+h) into the box.
-
You are not finished until the lone h in the denominator is
gone. An answer left as \tfrac{2xh + h^2}{h} is a formula
that self-destructs at exactly the value of h we care about:
plug in h = 0 and it reads \tfrac{0}{0}.
The entire purpose of the simplification is to cancel that h away —
legal while h \neq 0 — leaving an expression like
2x + h that stays meaningful as h
shrinks. If your "simplified" difference quotient still has a bare h
downstairs, the algebra has one more step in it. (A surviving x + h
downstairs, as in \tfrac{-1}{x(x+h)}, is fine — it tends to
x, not to 0.)
Fair question. Once you learn the derivative rules a few pages from now, you'll differentiate
x^2 in your head, and the expand-subtract-divide ritual can feel like
being made to churn butter when the shop sells it. But here's the secret: every rule in
the derivative toolbox was proved by someone doing exactly this simplification, once, in
general. The power rule is the difference quotient of x^n
simplified with the binomial expansion; the product rule is the difference quotient of
f(x)g(x) plus one sneaky added-and-subtracted term. When you do it
for x^2, you're not practising arithmetic — you're rehearsing the
proof technique behind the whole subject.
And the ritual is older than calculus itself. In the 1630s — decades before Newton or
Leibniz, with no notion of a limit anywhere in mathematics — Pierre de Fermat found maxima and
tangent lines by a method he called adequality: compare f(x)
with f(x + E), divide the difference by E,
simplify… and then simply discard whatever still contained E,
treating the step as zero after having divided by it as if it weren't. It worked beautifully and
it was logically indefensible, and everyone half-knew it: a century later the philosopher
Bishop Berkeley mocked these vanishing steps as "the ghosts of departed quantities."
The taunt stung because it was fair — and mathematics needed until the 19th century, and the
rigorous idea of a limit, to answer it properly. The algebra you just did is Fermat's method,
finally standing on solid ground: simplify first, so that letting h
vanish afterwards asks nothing illegal of the denominator.
Where this is pointing
You now hold the derivative's chrysalis. The difference quotient
\tfrac{f(x+h) - f(x)}{h} packages every average rate of
change near x into one expression with a shrink dial; simplifying it
removes the 0/0 booby-trap; and the shrinking-h
table shows the secant slopes converging on a single number — the curve's exact steepness at a
point. What remains is to make "let h shrink to nothing" an honest
mathematical operation, and to meet the line the secants tilt toward: the
tangent
line. When you get there, remember this page's ritual — build, subtract, divide,
simplify — it is the engine under everything that follows.
Watch it explained
Sal Khan finds the slope of a secant line for an arbitrary difference — exactly the
difference quotient, built and simplified on a live example. A good second pass over the
worked examples above.