From secant to tangent
On the curve $f(x) = x^2/2$, the point $P = (1, 0.5)$ is pinned and $Q$ sits a distance $h$ to the right. As $Q$ slides toward $P$, what does the slope of the line through $P$ and $Q$ head toward?
At $h = 1.80$ the readout shows $1.9000$. Slide down: $1.5000$ at $h = 1.00$, $1.2500$ at $h = 0.50$, $1.0500$ at $h = 0.10$, $1.0100$ at $h = 0.02$. The slope does not wander and does not blow up — it crowds in on $1$, and the pivoting secant settles onto the dashed line, the one the f′(1) readout has been holding the whole time.
The number you were watching is the difference quotient, $\frac{f(1+h) - f(1)}{h}$ — rise over run between $P$ and $Q$, the average rate of change over the interval. Setting $h = 0$ outright is illegal: the quotient reads $0/0$, and $P$ and $Q$ collapse into one point, which no longer determines a line. But the last chapter built the exact tool for this. Take the limit as $h \to 0$; here algebra finishes it in one line: $\frac{(1+h)^2/2 - 1/2}{h} = 1 + \frac{h}{2} \to 1$.
That limit is the slope of the curve at $P$ — the slope of the tangent line — and it earns a name last: the derivative of $f$ at $a$, written $f'(a)$.
Why is the slope at $P$ defined as a limit instead of by setting $h = 0$ in $\frac{f(1+h) - f(1)}{h}$?
The secant slope obeys the exact formula $m = 1 + h/2$. What must the readout show at $h = 0.50$, and what is $f'(1)$?
Transfer: for $g(x) = x^2$ at $a = 3$, the difference quotient simplifies: $\frac{(3+h)^2 - 9}{h} = 6 + h$. What is $g'(3)$?
No single secant is the tangent — the tangent is the limit of secants, and the derivative $f'(a)$ is its slope: an instantaneous rate of change, measured by shrinking an average.