The Derivative at a Point

Glance at a car's speedometer and it tells you, with total confidence: 60 km/h, right now. Not over the last minute, not since you left home — at this very instant.

Now think about what that actually claims. Speed is distance divided by time. To measure it you need the car to cover some distance in some time. But "right now" is a single instant: in an instant the car covers zero distance and zero time passes. Distance over time comes out as \tfrac{0}{0} — which is not a number at all. Taken literally, "speed at an instant" seems to be meaningless. Yet the needle sits there, pointing at 60.

This puzzle bothered thinkers for two thousand years, and the way out of it is one of the great ideas in all of mathematics. You can't divide zero by zero — but you can ask what the average speed is over a tenth of a second, then a hundredth, then a millionth, and watch what value those averages close in on. That target value — the limit — is what "60 km/h right now" really means. Today you'll learn its proper name: the derivative.

From average to instantaneous

You already know from the tangent line that a curve has a well-defined steepness at each point, found by sliding a secant line's second point toward the first. The slope of that secant is the difference quotient: step a distance h along from x = a, measure how much f rises, and divide rise by run:

\frac{\text{rise}}{\text{run}} = \frac{f(a + h) - f(a)}{h}.

For any actual step h \neq 0, this is an average rate of change — the speedometer averaged over a short stretch. The instantaneous rate is what these averages approach as the stretch shrinks to nothing. That number gets a name and a symbol: the derivative of f at a, written f'(a) and read "f prime of a".

Let f be defined on an interval around x = a. The derivative of f at a is f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}, provided this limit exists.
When the limit exists, f is called differentiable at a, and f'(a) is exactly the slope of the tangent line to y = f(x) at the point (a,\, f(a)).

Read the definition slowly — every piece earns its place. The fraction is the secant slope for a step of size h. The limit is the "close in on" machinery: it asks what single value the secant slopes approach as h shrinks — from the right (h > 0) and from the left (h < 0). And the whole thing is the exact mathematical answer to the speedometer paradox: instantaneous rate of change is not measured inside one instant, it is the limit of averages over ever-smaller windows around the instant.

Around 450 BC the Greek philosopher Zeno of Elea aimed a paradox at the very idea of motion. Watch an arrow in flight and freeze time at any single instant. At that instant the arrow occupies exactly one position — a space precisely its own length. It isn't moving within that instant (an instant has no duration for it to move in). But the arrow's whole flight is made of instants, and it's motionless at every one of them. So when, Zeno demanded, does it ever actually move?

For over two millennia there was no clean answer. The derivative finally supplies one: velocity at an instant was never a fact about the instant alone. It's a fact about the instant's neighbourhood — the limit of average velocities over shrinking windows around it. A single frozen frame of the arrow indeed contains no motion, just as \tfrac{0}{0} contains no answer; but the frames near it do, and the limit distils their information into one number. Zeno's frozen arrow and your speedometer are the same puzzle, and f'(a) resolves them both.

Sneak up on it: f'(2) for f(x) = x^2

Before doing any clever algebra, let's just watch the limit happen. Take f(x) = x^2 at a = 2, so f(2) = 4, and compute the difference quotient \frac{f(2+h) - 4}{h} for smaller and smaller steps:

h	2 + h	f(2+h) = (2+h)^2	\dfrac{f(2+h) - 4}{h}
1	3	9	5
0.5	2.5	6.25	4.5
0.1	2.1	4.41	4.1
0.01	2.01	4.0401	4.01
0.001	2.001	4.004001	4.001
-0.001	1.999	3.996001	3.999

Look at the last column. From above the values march 5,\ 4.5,\ 4.1,\ 4.01,\ 4.001; from below they climb 3.999, \ldots The quotients squeeze in on 4 from both sides. We never set h = 0 — every row uses an honest nonzero step — yet the target is unmistakable: f'(2) = 4.

The algebra confirms it, and explains the pattern in the table. With f(2+h) = (2+h)^2 = 4 + 4h + h^2:

\frac{(4 + 4h + h^2) - 4}{h} = \frac{4h + h^2}{h} = \frac{h(4 + h)}{h} = 4 + h.

There's the whole table in one expression! The quotient is exactly 4 + h: at h = 0.01 it reads 4.01, at h = -0.001 it reads 3.999. And once the troublesome h in the denominator has been cancelled, the limit is a simple substitution:

f'(2) = \lim_{h \to 0} (4 + h) = 4.

Watch the secant become the tangent

Here is that same computation as a picture. The point P = (2, 4) is fixed on y = x^2; the point Q sits a step h further along. The solid line through P and Q is the secant — its slope is the difference quotient 4 + h from the table above. Drag h toward zero and watch the secant settle onto the dashed tangent line of slope 4.

Notice the slider stops at h = 0.05, not at zero — because at h = 0 the points P and Q coincide and there is no secant line to draw. That is the geometric face of \tfrac{0}{0}. The tangent is not "the secant at h = 0"; it is the line the secants approach.

Worked example: f(x) = x^2 at a = 3

Now let's run the definition cleanly, start to finish, at a different point — every derivative-from-the-definition follows this same script. Find f'(3) for f(x) = x^2: build the difference quotient, simplify, then let the limit do its work. Start with f(3 + h) = (3 + h)^2 = 9 + 6h + h^2 and f(3) = 9:

\frac{(9 + 6h + h^2) - 9}{h} = \frac{6h + h^2}{h} = \frac{h(6 + h)}{h} = 6 + h

The 9s cancel, then a factor of h cancels — this is the crucial move, the step that dissolves the \tfrac{0}{0}. Now the denominator is gone, so we can find the limit by substitution — just put h = 0 into the simplified expression:

f'(3) = \lim_{h \to 0} (6 + h) = 6

So at x = 3 the curve y = x^2 has slope 6. The tangent line there rises six units for every one across. Compare with the previous example: at a = 2 we got 4, at a = 3 we get 6 — in both cases twice the value of a. Keep that pattern in mind; you'll see it again below.

Worked example: f(x) = x^3 at a = 1

A cube instead of a square — the script is identical, the expansion just has one more term. Find f'(1) for f(x) = x^3.

Step 1 — the two ingredients. We need f(1 + h) and f(1). Expanding the cube carefully (multiply (1+h)^2 = 1 + 2h + h^2 by (1+h) once more):

f(1 + h) = (1 + h)^3 = 1 + 3h + 3h^2 + h^3, \qquad f(1) = 1.

Step 2 — the difference quotient. Subtract and divide by h:

\frac{(1 + 3h + 3h^2 + h^3) - 1}{h} = \frac{3h + 3h^2 + h^3}{h} = \frac{h\,(3 + 3h + h^2)}{h} = 3 + 3h + h^2.

Step 3 — the limit. With h cancelled from the denominator, substitution is safe:

f'(1) = \lim_{h \to 0} \left( 3 + 3h + h^2 \right) = 3.

The tangent to y = x^3 at (1, 1) has slope exactly 3. Notice what survived the limit: only the term where h appeared to the first power in the expansion. The 3h^2 and h^3 terms still carried a factor of h after the cancellation, so they vanished as h \to 0. The derivative picks out the part of the change that is proportional to the step — everything of higher order melts away.

Three traps catch nearly everyone the first time through:

h \to 0 does not mean h = 0. If you set h = 0 straight away, the difference quotient reads \tfrac{f(a) - f(a)}{0} = \tfrac{0}{0} and you're stuck. The whole method is: simplify the quotient first (expand, cancel the factor of h), and only then take the limit. Throughout the algebra, h is a genuine nonzero number — that's exactly what makes dividing by it legal.
\tfrac{0}{0} is not an answer — it's a signal. It does not mean the derivative is 0, or 1, or undefined. It means "the interesting cancellation hasn't happened yet — keep working." Every derivative computed from the definition passes through a \tfrac{0}{0} disguise on its way to the true value.
f'(a) is a number, not a formula. f'(3) = 6 is a single slope at a single point of y = x^2. It says nothing (directly) about the slope at x = 5. A formula giving the slope at every point at once — like the 2a pattern you spotted above — is the next idea in the story, the derivative as a function; here we take the curve one point at a time.

See the tangent at any point

The definition hands you one slope per point — so let's collect a whole curve's worth. Slide the point a along f(x) = x^2. The bold tangent line always has slope f'(a) = 2a — the pattern our two worked answers (f'(2) = 4, f'(3) = 6) were hinting at: flat at the bottom (a = 0), steeper as you move out, and negative on the left, where the parabola runs downhill.

Watch the sign as you cross a = 0: the tangent tips from leaning left (negative slope) through perfectly flat to leaning right (positive slope). A derivative isn't just a steepness — its sign tells you whether the function is falling or rising at that point, and f'(a) = 0 flags the special flat spots.

The recipe, step by step

Every derivative-at-a-point you've seen — and every one you'll meet in an exam — follows the same four moves: form the difference quotient, expand the numerator, cancel the factor of h, then substitute h = 0 into what remains. Step through them on a fresh figure; the secant line from move 1 turns into the tangent at move 4.

Steps 2 and 3 are where the real work lives, and where the algebra differs from function to function — a square expands one way, a cube another, a reciprocal needs a common denominator. But steps 1 and 4 never change: the quotient going in, the substitution coming out.

Both — independently, and it caused one of the ugliest feuds in the history of science. Isaac Newton worked out his "method of fluxions" in the 1660s while sheltering from the plague, but, characteristically secretive, showed it to almost no one. Gottfried Wilhelm Leibniz developed the calculus independently in the 1670s and published first, in 1684, with the elegant \tfrac{dy}{dx} notation still used today. When the priority question exploded, the Royal Society convened an "impartial" committee — whose report was secretly drafted by its president, one Isaac Newton — and, shockingly, ruled for Newton. British mathematicians loyally stuck to Newton's clumsier notation for the next century and fell behind the Continent as a result. Today both men share the credit, and you're using a third inventor's shorthand anyway: the tidy f'(a) "prime" notation was introduced later by Joseph-Louis Lagrange.

Watch it explained

Sal Khan calculates the slope of a tangent line at a point using the definition of the derivative — the same four-move recipe, narrated over a worked example.