Drop a ball and ask “how fast was it going?” Between two instants, the answer is easy: total distance over total time. On a graph of position, that’s exactly the slope of the line connecting the two points — the secant line. This is the average rate of change, and you already know it.
What you're looking at: the curve is \(f(x)=x^2\). The line through the two dots is the secant; its slope is the average rate of change of \(f\) between \(a\) and \(b\).
This is the slope of the secant line. It's an average — it smooths over everything that happens between the two points.
Average speed over two seconds isn't the speed right now. To get the rate at a single instant, slide the second point toward the first and watch the secant pivot. As the gap closes, the secant settles onto one specific line — the tangent — and its slope settles on one number. That limiting slope is the instantaneous rate of change.
What you're looking at: the fixed point sits at \(x=1\). The second point is a distance \(h\) away. As you push \(h\) toward \(0\), the secant pivots into the tangent at \(x=1\).
The secant slope never reaches a fixed value by arithmetic alone — but it's clearly heading toward 2. That limit is the slope of the tangent, and the speed at the instant \(x=1\).
Let's make “the limit of the secant slope” precise. The secant runs from \(x=a\) to a point a distance \(h\) away, so its slope is \(\frac{f(a+h)-f(a)}{h}\). The derivative is what that slope approaches as \(h\to 0\). That's the whole definition:
When this limit exists, we call its value \(f'(a)\) — “f prime of a” — the derivative of \(f\) at \(a\). It is a number: the slope of the tangent line at that point.
Here it is in action — no shortcut rules, just the definition and the Unit 1 algebra for killing a 0/0. Step through finding the derivative of \(f(x)=x^2\):
Setting \(h=0\) at the start gives \(\frac{0}{0}\) — undefined. So we do the algebra first: expand, simplify, and cancel the \(h\) in the denominator. Only once the \(h\) downstairs is gone can we let \(h\to 0\) safely. This is exactly the “\(0/0\) means do more work” move from Unit 1.
Mathematicians invented several notations for the derivative, and the AP exam uses all of them interchangeably. They mean exactly the same object — the rate of change of \(y=f(x)\) — so don't let the different symbols rattle you.
\(f'(x)\) is a function — it gives a slope for every \(x\). Plug in a specific input and \(f'(a)\) is a number — the one slope at that point. Same idea as \(f(x)\) vs. \(f(3)\).
The derivative turns one function into another: feed it \(x^2\), it hands back \(2x\), a brand-new function whose output is the original's slope.
On the AP exam you're often handed a function only as a table or a graph — no formula to differentiate. You can still estimate the derivative: it's a slope, so use the nearest points you have. This exact skill shows up on a free-response table question almost every year.
Pick the two table points straddling your target and take rise over run — the symmetric difference \(\frac{f(x+h)-f(x-h)}{2h}\). It's the secant slope across the gap, and it's the best estimate the table allows.
At an endpoint, you only have one neighbor, so use a one-sided secant. Always show the difference quotient — that's where the AP points live.
Watch the graph: the gold line is the secant your estimate uses. Its slope is the number you compute — estimating \(f'\) from a table is just measuring that line.
Strip away the words and a derivative is just a slope at a point — rise over run, where the run has shrunk to nothing. Every later rule is a shortcut for finding it.
The secant slope over an interval turns into the tangent slope at a point when you take the limit as the gap → 0. That limit is the entire definition.
The difference quotient is \(\frac00\) if you plug in \(h=0\) too early. Simplify first, cancel the \(h\), then take the limit — the Unit 1 move.
\(f'(x)\), \(\frac{dy}{dx}\), \(\frac{d}{dx}f\) all name the same derivative. Different notations highlight different things, but the object is one.
Five problems across the lesson — average rates, the limit definition, table estimates, and tangent slopes. Try each before revealing.
Eight fast checks across the whole lesson.
The average rate over \([a,b]\) is a secant slope; the instantaneous rate at \(a\) is the tangent slope (a limit). A question asking for the rate “at \(t=3\)” wants the derivative, not a secant.
Plugging \(h=0\) into the difference quotient gives \(\frac00\). You must expand and cancel the \(h\) in the denominator first, then take the limit.
Until the \(h\) is cancelled, every line still needs \(\lim_{h\to0}\). On the AP exam, an answer that loses the limit notation mid-derivation can lose points.
An instantaneous rate of change carries units: feet per second, dollars per item. A bare number with no interpretation often misses the point of the question.
The value of the function and the value of its derivative are different things. \(f(a)\) is the height; \(f'(a)\) is the slope there.
The AP table question awards points for the setup \(\frac{f(x_2)-f(x_1)}{x_2-x_1}\), not just the final number. Always write it out.