The derivative as a limit

A slope needs two points. That is not a limitation of the notation — it is what a slope is: a change in one thing divided by a change in another. Ask for the slope at a single point and the question appears to be malformed, because dividing by a change of zero is not a small number, it is nothing at all.

And yet “how fast is this changing right now” is an obviously sensible question. A car has a speed at the instant you glance at the speedometer, not just an average over the last minute.

The whole of differential calculus is the trick for answering it. Take a second point, at a distance h. Compute the slope you can actually compute. Then let h go to zero and watch where the answer settles.

The line through the two points is a secant. As h shrinks, the second point slides towards the first, the secant pivots, and its slope approaches something. At h = 1.5 the slope reads 3.5; at h = 0.5, 2.5; at h = 0.01, 2.01. It is heading for 2, and it never gets there, because at h = 0 there is no second point and no line.

That is the point worth sitting with: the answer is not the value at zero. It is the value the answer is heading towards.

The difference quotient

For a function $f$ and a point $a$, the slope of the secant through $(a,\,f(a))$ and $(a+h,\,f(a+h))$ is

$$\frac{f(a + h) - f(a)}{h}$$

and the derivative is what that approaches as $h \to 0$:

$$f'(a) \;=\; \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$

Every derivative rule you will meet is a shortcut for this limit. They are worth knowing, and they are not worth mistaking for the definition — a rule tells you the answer, and the definition tells you what the answer means.

Doing it by hand, once

Take $f(x) = x^2$, at $a = 1$. Expand:

$$\frac{(1+h)^2 - 1^2}{h} \;=\; \frac{1 + 2h + h^2 - 1}{h} \;=\; \frac{2h + h^2}{h} \;=\; 2 + h$$

That last cancellation is the entire trick. Before it, the expression is $\tfrac{0}{0}$ at $h = 0$ and tells you nothing. After it, the expression is $2 + h$, which is perfectly well behaved at $h = 0$ and says $2$.

The two expressions are equal everywhere except at $h = 0$, and the limit only cares about what happens near zero, not at it. That is why the cancellation is legitimate.

The third line is the general result: the slope at $a$ is $2a$, which at $a = 1$ is $2$ — the number the widget above was creeping towards.

Why the limit is not optional

It is tempting to read the difference quotient as “the slope when h is very small” and skip the limit. Two things go wrong.

The first is that “very small” is not a number. Any $h$ you pick gives a secant, not a tangent, and the answer it gives is wrong by an amount you have chosen not to measure. For $x^2$ at $1$ the secant slope is exactly $2 + h$ — so picking $h = 0.001$ gives $2.001$, and the error is not a rounding artefact but the actual answer to a slightly different question.

The second is that the limit can exist when no single small value is right. That is the case the definition handles and arithmetic does not.

Those three are the derivatives of $\sin$, $\cos$ and $\exp$ at zero, each computed straight from the definition, and none of them simplifies by cancelling a factor of $h$. There is no algebra that turns $\tfrac{\sin h}{h}$ into something defined at zero — the limit is the only route to the answer.

Where it fails

A derivative is a limit, and a limit need not exist. Three ways a function can have no derivative at a point:

• A corner. $|x|$ at $0$: approach from the right and the secant slope is $1$, from the left it is $-1$. The two one-sided limits disagree, so there is no limit and no derivative — even though the function is perfectly continuous.
• A vertical tangent. $\sqrt[3]{x}$ at $0$: the secant slope grows without bound. The tangent line exists geometrically; its slope does not exist as a number.
• A discontinuity. If the function jumps, the difference quotient does not settle. Differentiability implies continuity, and this is the contrapositive.

The first is the one worth remembering, because it separates two ideas students routinely fuse: continuous does not mean smooth. $|x|$ has no break in it and still has no derivative at the origin.

That last line is deliberately vacuous, and it is in the chapter to make a point about this book’s own machinery: the checker will confirm an identity that says nothing. A verified claim is not automatically a meaningful one. What fails for $|x|$ is not an identity but the existence of a two-sided limit, and that is a fact about limits rather than an equation to check.