The derivative as a function AB & BC

Replace the fixed $a$ with a free variable $x$ and the limit from Topic 2.1 no longer produces a number. It produces a rule:

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

This is the derivative function. Its input is a point, and its output is the slope of the tangent to $f$ there. The domain of $f'$ is every $x$ at which that limit exists, which can be smaller than the domain of $f$ (Topic 2.4 is about exactly where it shrinks).

The two objects are related in one direction only. Once $f'(x)$ is known, evaluating at a point gives $f'(a)$ for free. Computing $f'(a)$ at a single $a$ tells you nothing about $f'$ anywhere else.

A function is differentiable at $a$ when that limit exists at $a$, and differentiable on an interval when it exists at every point of the interval.

Several notations coexist because they emphasize different things. They all name one of two objects: a function, or a number.

1. $f'(x)$ and $\frac{dy}{dx}$ and $\frac{d}{dx}\left[f(x)\right]$ are functions. Lagrange's prime is compact; Leibniz's $\frac{dy}{dx}$ keeps the input variable visible, which matters the moment there is more than one.
2. $f'(a)$ and $\left.\frac{dy}{dx}\right|_{x = a}$ and $y'(a)$ are numbers: the derivative function evaluated at $a$.
3. $\frac{d}{dx}$ on its own is an instruction, not a quantity. It has to be applied to something.

The vertical bar in $\left.\frac{dy}{dx}\right|_{x=2}$ is the Leibniz way of writing "and now evaluate at 2". Without it, $\frac{dy}{dx}$ is still a whole function, so an answer of $\frac{dy}{dx} = 6$ when a specific point was requested is answering a different question.

$\frac{dy}{dx}$ is written like a fraction and behaves like one in a few places you will meet later, but it is defined as a limit, not as a ratio of two numbers called $dy$ and $dx$. Splitting it apart is not a legal move in this unit.

The procedure is fixed: expand, subtract, cancel the $h$, take the limit. Cancelling is legal because $h$ is never zero during the limit process.

Take $f(x) = x^2 - 4x$:

$$f(x + h) - f(x) = \left[(x+h)^2 - 4(x+h)\right] - \left[x^2 - 4x\right] = 2xh + h^2 - 4h.$$

$$\frac{2xh + h^2 - 4h}{h} = 2x + h - 4 \longrightarrow f'(x) = 2x - 4.$$

Now $f'(3) = 2$ and $f'(0) = -4$ come out by substitution, with no new limit. That is the whole payoff of computing the function rather than the point.

Not every case cancels by expanding. For $f(x) = \frac{1}{x}$ the numerator has to be combined over a common denominator first:

$$\frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \frac{\frac{-h}{x(x+h)}}{h} = \frac{-1}{x(x+h)} \longrightarrow f'(x) = -\frac{1}{x^2}.$$

For a root, multiply by the conjugate instead. Both are Topic 1.6 techniques applied to a quotient that always starts as $\frac{0}{0}$.

Because $f'$ is a function, its graph can be drawn, and the two graphs answer different questions. The height of $f$ at $x$ is a value; the height of $f'$ at $x$ is a slope of $f$. Where $f$ has a horizontal tangent, $f'$ crosses zero. Where $f$ is falling, $f'$ is negative.

In an applied setting the units follow the same rule as in Topic 2.1: output units of $f$ per input unit. If $C(q)$ is the cost in dollars of producing $q$ items, then $C'(q)$ is in dollars per item, and $C'(50) = 12$ means the cost is rising at about 12 dollars per item at a production level of 50.

Notation carries that meaning. $\frac{dC}{dq}$ says at a glance which quantity is changing with respect to which, and it is the reason Leibniz notation dominates in applications.