Answers · The mathematics underneath

How can there be a slope at a single point if slope needs two points?

In shortA slope at one point is the single value that slopes between nearby pairs of points settle on as the pair closes up; it is not a slope measured between two identical points.

There can be a slope at a single point because the slope at a point is not computed from one point alone. It is the number that the slopes between two nearby points settle on as the second point is moved towards the first. Write the rule as $f$ , the fixed input as $x$ , and the gap to a second input as $h$ . Across the interval from $x$ to $x+h$ the average slope is the rise divided by the run, and the run is $h$ . The instantaneous slope is the value that these averages approach as $h$ approaches zero.

f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.

What the limit is doing

The $h$ in the fraction is never zero. That matters because division by zero is undefined: if $h$ were zero, both the rise $f(x+h)-f(x)$ and the run $h$ would be zero, and the expression would be $0/0$ . The limit is not a command to put $h=0$ into that fraction. It is a statement about the numbers the fraction produces for small nonzero $h$ : they can be made as close as anybody likes to one fixed value, and they stay close. When that happens, the fixed value is called $f'(x)$ .

FIG 1The slope of the squaring rule near an input of two

step	width of the interval	rise across it	rise divided by width	what happened
1	1	5	5	From two to three, the answer rose by five.
2	0.5	2.25	4.5	Half the interval, and the quotient has dropped to 4.5.
3	0.25	1.0625	4.25	A quarter of the interval gives 4.25.
4	0.1	0.41	4.1	The quotient is four plus the width.
5	0.01	0.0401	4.01	Still four plus the width; the settling is calm.

5 steps

At an input of two the squaring rule gives four. Each row takes a second input at a distance to the right, forms the rise, and divides by . The quotient is at every row, so the values are heading for four.

Take $f(x)=x^2$ and ask for the slope at $x=2$ . At a gap of $h$ the second input is $2+h$ , and the second answer is $(2+h)^2$ . The rise is $(2+h)^2-4$ . Multiply out: $(2+h)^2=4+4h+h^2$ , so the rise is $4h+h^2$ . Divide by the run $h$ : the average slope is $(4h+h^2)/h=4+h$ .

\frac{(2+h)^2-2^2}{h} = \frac{4+4h+h^2-4}{h} = \frac{4h+h^2}{h} = 4+h \to 4 \quad \text{as } h\to0.

Geometrically, the fraction is the slope of a straight line through two points on the curve: $(x,f(x))$ and $(x+h,f(x+h))$ . As $h$ shrinks, the second point slides towards the first, and the line pivots about the fixed point. The limiting line is the tangent, and the tangent's slope is $f'(x)$ . Every intermediate calculation used two points; the answer is still the slope at one point because it is the value those two-point slopes settle on.

Why the obvious calculation fails

The natural guess is to make the two points identical and compute rise over run directly. Then the rise is $f(x)-f(x)=0$ and the run is $x-x=0$ , so the formula gives $0/0$ . That is not a slope; it is the failure of the direct formula. The definition avoids the failure by doing the division while $h$ is still nonzero. In the squaring example the division gives $4+h$ exactly, and only after that cancellation is it safe to let $h$ go to zero. Setting $h=0$ before cancelling would leave $0/0$ and say nothing. Setting it after cancelling gives $4$ .

For a car, average speed over an interval is distance divided by time. The instantaneous speed is the value those averages settle on as the interval shrinks. The speedometer is reporting that settling value, not a distance travelled in zero time.

The definition is local. It uses inputs arbitrarily close to $x$ , but it never uses an input equal to $x$ on both sides of the division. That is why a rule can have a perfectly definite slope at one input and a different slope at another, and why no knowledge of the rule far away is needed.

Where there is no slope

The settling is not guaranteed. Take $f(x)=|x|$ at $x=0$ . From the right, with $h>0$ , the quotient is $(|h|-0)/h=h/h=1$ . From the left, with $h<0$ , it is $(|h|-0)/h=(-h)/h=-1$ . The two one-sided values stay one apart no matter how small $|h|$ is, so there is no single number for the quotients to settle on, and no slope exists at the corner. A sharp bend in a graph, a switch in a signal, or a change of band in a cost can produce the same failure. Where the left and right quotients disagree, or where they grow without bound, the instantaneous slope is not defined.

Worth keeping

A slope at a point is defined as the limit of average slopes over shrinking intervals, not as a division at a single isolated input.
Before setting the interval width to zero, the algebra must cancel it; otherwise the expression is zero divided by zero.
At a corner the slopes from the left and right disagree, so no single instantaneous slope exists there.

Where this goes next

This question is one step of Change and Accumulation, a course that works through its subject from the beginning. The page that teaches this part of it is How Fast, Exactly Here, and it reads the whole way through whether or not you have an account.

What an account changes is the practice: problems standing under every idea, marked with a reason rather than a tick, in a copy of the course that is yours.

Read that page The whole course