In AP Calculus, the slope of the tangent line to a curve at a point equals the derivative of the function at that point, f'(a), and represents the instantaneous rate of change there. It is defined as the limit of secant line slopes (average rates of change) as the interval shrinks to zero.
The slope of the tangent line is the geometric face of the derivative. If you zoom in on a smooth curve at a single point, the curve starts to look like a straight line. That line is the tangent line, and its slope tells you exactly how steep the curve is, and how fast it's changing, at that one instant.
Here's the catch that makes this a calculus idea and not an algebra idea. Slope needs two points (rise over run, Δy/Δx), but a tangent line touches the curve at just one point. So you can't compute it directly. Instead, you take secant lines through two nearby points and let the second point slide toward the first. The secant slopes are average rates of change, and their limit as the gap h goes to 0 is the tangent slope. That limit, lim(h → 0) (f(x+h) - f(x))/h, is the derivative. So "slope of the tangent line at x = a" and "f'(a)" are two names for the same number.
This idea is the bridge between Unit 1 (Limits and Continuity) and Unit 2 (Differentiation). Topic 1.1 asks the founding question of the course, can change occur at an instant? The answer (per EK CHA-1.A.3) is yes, if you define instantaneous rate of change as a limit of average rates of change. Topic 2.2 then makes it official. Learning objective 2.2.A has you represent the derivative as the limit of a difference quotient, and 2.2.B has you find the equation of a tangent line at a given point, using the essential knowledge that the derivative at a point IS the tangent slope. Almost everything later in the course (linear approximation, motion, related rates, optimization) is built on this one equivalence, so if you can say "derivative = tangent slope = instantaneous rate of change" in your sleep, you're set up for the whole exam.
Keep studying AP Calculus Unit 5
Visual cheatsheet
view galleryDerivative / f'(x) (Unit 2)
These aren't just related, they're the same thing viewed two ways. The derivative is the analytic answer (a limit you compute) and the tangent slope is the graphical answer (a steepness you can see). The CED literally defines the derivative at a point as the slope of the tangent line there.
Average Rate of Change (Secant Line) (Unit 1)
A secant line connects two points on the curve, so its slope is an average rate of change over an interval. The tangent slope is what those secant slopes approach as the two points merge. Secant is the approximation, tangent is the limit.
Difference Quotient (Unit 2)
(f(x+h) - f(x))/h is just the slope formula Δy/Δx written for two points on the curve that are h apart. Take the limit as h → 0 and the difference quotient becomes the tangent slope. It's rise-over-run upgraded with a limit.
Local Extrema (Unit 5)
At a smooth peak or valley, the tangent line goes flat, so its slope is zero. That's why setting f'(x) = 0 finds candidates for local maximums and minimums. The tangent slope idea from Unit 2 quietly powers all of Unit 5's analysis.
Multiple choice loves this term as a translation test. A stem gives you "the slope of the tangent line to a curve at a point" and the right answer is the derivative at that point, or the instantaneous rate of change, or the limit of the difference quotient. Practice questions hit it from all four directions, asking what the tangent slope represents, which rates of change require limits, and how to read the derivative off a graph. You also need to compute with it. Topic 2.2.B questions hand you a point and ask for the tangent line equation, which means finding f'(a) for the slope and plugging into y - f(a) = f'(a)(x - a). On FRQs, tangent line equations show up as a standard sub-part, and "interpret f'(c) in context with units" questions are really asking you to read a tangent slope as a rate.
The secant slope uses two distinct points and measures the average rate of change over an interval. The tangent slope lives at one point and measures the instantaneous rate of change. You can compute a secant slope with plain algebra (Δy/Δx), but the tangent slope requires a limit, because plugging in a single point makes the denominator zero (EK CHA-1.A.2). On the exam, "over the interval [a, b]" signals secant, while "at x = a" or "at the instant" signals tangent.
The slope of the tangent line at a point equals the derivative of the function at that point, f'(a).
Tangent slope, derivative at a point, and instantaneous rate of change are three names for the same number, and the exam swaps between them freely.
You can't compute a tangent slope with the plain slope formula because one point gives a zero denominator, so calculus defines it as the limit of secant slopes as h → 0.
The defining limit is lim(h → 0) (f(x+h) - f(x))/h, which is the difference quotient with the gap between points shrinking to zero.
To write a tangent line equation, evaluate f'(a) for the slope, then use point-slope form: y - f(a) = f'(a)(x - a).
A horizontal tangent line means the derivative is zero, which is how you find candidates for local extrema in Unit 5.
It's the steepness of a curve at a single point, equal to the derivative of the function evaluated there. Formally, it's the limit of the difference quotient, lim(h → 0) (f(x+h) - f(x))/h, which is the limit of secant slopes as the two points merge into one.
Yes, at a point they're the same number. The CED's essential knowledge for 2.2.B states that the derivative of a function at a point is the slope of the tangent line to the graph at that point. The derivative f'(x) is the function that outputs tangent slopes for every x.
A secant line passes through two points on the curve and its slope is an average rate of change over that interval. A tangent line touches at one point and its slope is the instantaneous rate of change. The tangent slope is defined as the limit of secant slopes as the interval shrinks to zero.
Two ingredients. First find the point (a, f(a)), then compute the slope f'(a). Plug both into point-slope form, y - f(a) = f'(a)(x - a). This is exactly what learning objective 2.2.B asks you to do, and you can leave the answer in point-slope form on the exam.
Because slope is change in y divided by change in x, and at a single point the change in x is zero, making the formula undefined (EK CHA-1.A.2). The limit lets you sneak up on the answer by computing secant slopes over smaller and smaller intervals and seeing what value they approach.