Fiveable

♾️AP Calculus AB/BC Unit 2 Review

QR code for AP Calculus AB/BC practice questions

2.2 Defining the Derivative of a Function and Using Derivative Notation

♾️AP Calculus AB/BC
Unit 2 Review

2.2 Defining the Derivative of a Function and Using Derivative Notation

Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 examWritten by the Fiveable Content Team • Last updated September 2025
♾️AP Calculus AB/BC
Unit & Topic Study Guides
Pep mascot

AI

Welcome back to AP Calculus with Fiveable! This topic focuses on determining derivative notation and calculating the derivative of a function. Let's combine our skills in calculating rates of change with the knowledge of limits to continue building our derivative skills. 🧱

Pep mascot
more resources to help you study

↗️ Definition of Derivative

The derivative of a function at a single point is the instantaneous rate of change at that point. We learned how to calculate an instantaneous rate of change in the previous topic: Defining Average and Instantaneous Rates of Change at a Point.

But how can we find the derivative of the whole curve, instead of at just one single point? It would be far too tedious to calculate the instantaneous rate of change at every single point, and then graph. 🤔

We can actually find the derivative by generalizing the limit notation and not solving for the derivative at the point. Therefore, we use the limit definition of a derivative:

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

You will learn certain rules and shortcuts to calculating derivatives in the next few lessons, but for now, we will use the definition of the derivative to calculate the rate of change for a function.

💡 Since the derivative represents a rate of change, the derivative at a point will represent the slope of a line tangent to the curve at that point! Check it out in the graph below.

Screen Shot 2023-12-19 at 6.34.43 PM.png
Graph of a tangent line, created with Desmos

We can see that the slope of the tangent line is f(1)f'(1), or the derivative of f(x)f(x) when x=1x = 1.

📝 Derivative Notation

There are several ways to represent a derivative in calculus.

If the original function can be represented as y=f(x)y = f(x), then the derivative can be represented as yy', f(x)f'(x), or dydx\frac {dy}{dx}. These all describe the rate of change of the function as the dependent variable changes.

To reiterate,

y=f(x)=dydxy' = f'(x) = \frac{dy}{dx}

They are all valid, and mean the same thing!


🧮 Practice Problems

Let’s work on a few questions and make sure we have the concept down!

  1. Using the Definition of a Derivative

Given y=3x2+4xy = 3x^2 + 4x, calculate yy'.

Let’s plug in all of the given information to the limit. Since y=f(x)y = f(x),

y=limh0[3(x+h)2+4(x+h)][3x2+4x]hy' = \lim_{h \to 0} \frac{[3(x + h)^2 + 4(x+h)]- [3x^2+4x]}{h}

Let's simplify this so that we can take the limit and determine how the function changes. We can begin by expanding the numerator and then distributing it.

y=limh03(x2+2xh+h2)+4(x+h)(3x2+4x)hy'= \lim_{{h \to 0}} \frac{{3(x^2 + 2xh + h^2) + 4(x+h)-(3x^2 + 4x)}}{h}

y=limh03x2+6xh+3h2+4x+4h3x24xhy'= \lim_{{h \to 0}} \frac{{3x^2 + 6xh + 3h^2 + 4x + 4h - 3x^2 -4x}}{h}

Now, let’s combine like terms.

y=limh06xh+3h2+4hhy'= \lim_{{h \to 0}} \frac{{6xh + 3h^2 + 4h}}{h}

We are almost there! Since h0h \neq 0, we can divide by 00.

y=limh0(6x+3h+4)y' = \lim_{{h \to 0}} (6x + 3h + 4)

As hh approaches 00, the middle term will approach 00. Therefore, we conclude:

y=6x+4y' = 6x + 4

Amazing work! 🎉

2) Tangent Line to a Curve

Given the curve f(x)=1xf(x) = \frac {1}{x}, find the equation of the line tangent to the curve at (1,1)(1,1).

Let’s use the definition of derivative to first calculate the derivative at (1,1)(1,1).

f(x)=limh01x+h1xhf'(x) = \lim_{{h \to 0}} \frac{\frac{1}{{x + h}} - \frac{1}{x}}{h}

This might seem a bit intimidating, but here we can change the two fractions in the numerator to have a common denominator.

In the numerator, we then have x(x+h)x(x+h)\frac{x - (x + h)}{x(x + h)}. Therefore…

f(x)=limh0x(x+h)x(x+h)hf'(x) = \lim_{{h \to 0}} \frac{\frac{x - (x + h)}{x(x + h)}}{h}

Now, we can multiply the numerator by the reciprocal of the denominator.

f(x)=limh0x(x+h)x(x+h)1h=f(x)=limh0x(x+h)x(x+h)(h)f'(x) = \lim_{{h \to 0}} {\frac{x - (x + h)}{x(x + h)}}{\frac{1}{h}} = f'(x) = \lim_{{h \to 0}} {\frac{x - (x + h)}{x(x + h)(h)}}

Now let’s expand the numerator.

f(x)=limh0xxhx(x+h)(h)=limh0hx(x+h)(h)f'(x) = \lim_{{h \to 0}} {\frac{x - x - h}{x(x + h)(h)}}= \lim_{{h \to 0}} {\frac{- h}{x(x + h)(h)}}

We can cancel the hh from both the numerator and the denominator to get

f(x)=limh01x(x+h)=limh01x2+xhf'(x) = \lim_{{h \to 0}} \frac{-1}{x (x + h)} = \lim_{{h \to 0}} \frac{-1}{x^2 + xh }

As hh approaches 00, the second term in the denominator will approach 00, so we can conclude

f(x)=1x2f'(x)= \frac{-1}{x^2}

Now we need to write the equation of a tangent line! Remember that a line can be represented in point-slope form as the following: yy1=m(xx1)y-y₁=m(x-x₁) where mm is the slope, or the derivative at the point (x1,y1)(x_1,y_1).

Our point is (1,1)(1,1), so f(1)=1(1)2=1f'(1)= \frac{-1}{(1)^2} = -1.

Now we have all of the necessary information to write the equation of the tangent line:

y1=1(x1)y-1=-1(x-1)

Great work! Let's check that this line is tangent to the curve $f(x)= \frac{1}{x}$ at the point $(1,1)$.

Screen Shot 2023-12-19 at 6.54.08 PM.png
Graph of tangent line, created with Desmos

Looks great! 🙌


🌟 Closing

Great job! 🚀👩‍🚀 You're mastering these concepts, and with practice, you'll navigate derivatives and continuity with confidence. Determining derivatives and using different derivative notations are crucial skills in AP Calculus. As you encounter questions on the exam, remember to check for domain restrictions and assess piecewise continuity.

Frequently Asked Questions

What's the formula for finding the derivative using limits?

The derivative of a function is defined as a limit of a difference quotient. The two common forms are: - At a general x: f′(x) = lim_{h→0} [f(x+h) − f(x)] / h (if this limit exists)—this gives the derivative function. - At a specific point a: f′(a) = lim_{x→a} [f(x) − f(a)] / (x − a) (equivalent form). Notation you’ll see on the AP: f′(x), dy/dx, or y′. Conceptually, this limit gives the instantaneous rate of change of f at x and is the slope of the tangent line there (useful for CHA-2.B and CHA-2.C tasks on the AP). For tangent-line equation at x = a: y = f(a) + f′(a)(x − a). For more AP-aligned review, see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and try practice problems (https://library.fiveable.me/practice/ap-calculus).

How do I solve the limit definition of a derivative step by step?

Use the limit definition: f'(x) = lim_{h→0} [f(x+h) − f(x)]/h. Step-by-step for a concrete f (say f(x)=x^2): 1. Plug into the difference quotient: [ (x+h)^2 − x^2 ] / h. 2. Expand and simplify numerator: (x^2 + 2xh + h^2 − x^2)/h = (2xh + h^2)/h. 3. Factor h: h(2x + h)/h. 4. Cancel h (for h ≠ 0): 2x + h. 5. Take the limit as h → 0: lim_{h→0} (2x + h) = 2x. So f'(x)=2x. Key tips tied to the CED: always simplify algebraically before taking the limit (CHA-2.B.2). If the left and right limits as h→0 disagree, the derivative doesn't exist (differentiability at a point, CHA-2.C.1). The derivative at x gives the tangent slope—use it to write the tangent line: y = f(a) + f'(a)(x−a). For more worked examples and AP-style practice, see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and the Unit 2 overview (https://library.fiveable.me/ap-calculus/unit-2). For extra practice, try problems at (https://library.fiveable.me/practice/ap-calculus).

When do I use f'(x) vs dy/dx vs y' notation?

Use whichever notation fits the context—they all mean “the derivative,” but each is more convenient in different situations (CED CHA-2.B.3). - f′(x): best when you think of the derivative as a function of x (analytic work, applying derivative rules, writing f′(x)=limit of the difference quotient). Clean for formulas and graph/analysis tasks (CHA-2.B.2, CHA-2.B.4). - dy/dx: Leibniz form is great when variables and rates matter (related rates, differentials, separation of variables, or emphasizing “rate of change of y with respect to x”). It makes chain rule and implicit differentiation look natural. - y′ (or f′): short and handy when y = f(x) and you’re working pointwise (tangent-line slope at a point) or doing quick algebraic steps. Examples: slope of tangent at x = a → use f′(a) or (dy/dx)|_{x=a}; implicit differentiation → use dy/dx; concise manipulation in proofs → y′ or f′(x). For higher derivatives use f′′(x) or d²y/dx² as appropriate. For review and practice, see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and Unit 2 overview (https://library.fiveable.me/ap-calculus/unit-2). For extra practice problems, go to (https://library.fiveable.me/practice/ap-calculus).

What's the difference between the derivative at a point and the derivative function?

The derivative at a point is a single number: the limit of the difference quotient at x = a, f′(a) = lim_{h→0} [f(a+h)-f(a)]/h (when it exists). It’s the instantaneous rate of change there and the slope of the tangent line at that one x-value. The derivative function, written f′(x) or dy/dx, is a new function that gives those pointwise slopes for every x where the limit exists (CHA-2.B.2, CHA-2.C.1). So: f′(2) is one slope; f′(x) is a rule you can evaluate to get slopes at any x. If f′(a) doesn’t exist, the derivative function is undefined at a (non-differentiability). This distinction appears often on the AP exam—be ready to use the limit definition, graph/table/analytic representations, and to write tangent-line equations (see Topic 2.2 study guide: https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD; unit overview: https://library.fiveable.me/ap-calculus/unit-2). For extra practice, try problems at (https://library.fiveable.me/practice/ap-calculus).

I don't understand how to set up the difference quotient - can someone explain?

The difference quotient is just the slope of a secant line through (x, f(x)) and (x+h, f(x+h)). Set it up as (f(x+h) − f(x)) / h, then take the limit as h → 0 to get the derivative f′(x) (CHA-2.B.2). Steps to use it: 1. Write f(x+h). 2. Form f(x+h) − f(x). 3. Divide that difference by h. 4. Algebraically simplify (cancel h factors). 5. Take the limit as h → 0. Quick example: f(x)=x^2. - f(x+h)=(x+h)^2 = x^2 + 2xh + h^2 - Difference: (x^2+2xh+h^2) − x^2 = 2xh + h^2 - Divide by h: (2xh + h^2)/h = 2x + h - Limit as h→0: f′(x)=2x Remember: if you can’t cancel h, the limit may not exist (not differentiable). For AP review and more examples see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and thousands of practice problems (https://library.fiveable.me/practice/ap-calculus).

How do I find the equation of a tangent line at a specific point?

1) Get the point: you need (a, f(a)). 2) Find the slope of the tangent = f′(a). Either - Use the limit (difference quotient): f′(a) = lim_{h→0} [f(a+h)−f(a)]/h (CED CHA-2.B), or - Use derivative rules (power, product, chain, etc.) and then evaluate at x = a (CED CHA-2.B.3). 3) Write the line in point-slope form: y − f(a) = f′(a)(x − a). That’s the tangent line. Quick notes: if you have parametric equations, slope = (dy/dt)/(dx/dt) at the given t. If f′(a) doesn’t exist, there’s no tangent line there (not differentiable). On the AP exam show work: either the limit computation or the derivative rule you used and the evaluation, then the point-slope step. For more examples and step-by-step practice, see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and extra problems (https://library.fiveable.me/practice/ap-calculus).

Why does the limit as h approaches 0 give me the slope of the tangent line?

Think of the difference quotient, (f(x+h) − f(x))/h, as the slope of a secant line through the points (x, f(x)) and (x+h, f(x+h)). As h gets smaller those two points get closer, so the secant line “rotates” and approaches one limiting line—the tangent at x. The limit as h → 0 picks out that limiting slope (if it exists), which is the instantaneous rate of change of f at x. That limiting value is the derivative f′(x) (CHA-2.B.2) and, by definition, is the slope of the tangent line at that point (CHA-2.C.1). Geometrically: secants → tangent; analytically: difference quotients → derivative. For more practice and the AP framing, see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) or Unit 2 overview (https://library.fiveable.me/ap-calculus/unit-2).

What does it mean when they say the derivative doesn't exist at a point?

Saying “the derivative doesn’t exist at a point” means the limit that defines the derivative, lim_{h→0} [f(x+h)-f(x)]/h, does not exist (so f′(x) is not a real number). That can happen for a few AP-tested reasons: the two one-sided difference-quotient limits aren’t equal (a corner), one or both blow up to ±∞ (a vertical tangent), or f isn’t even continuous there (so the derivative can’t exist). Graphically, there’s no single tangent line with a well-defined slope at that x. On the exam you’ll often justify nondifferentiability by showing the limit fails or by pointing to discontinuity, corner, or vertical tangent (CED CHA-2.B and CHA-2.C). For a quick review, see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and try practice problems at (https://library.fiveable.me/practice/ap-calculus).

How do I use the limit definition to find the derivative of x^2?

Start with the limit definition (CHA-2.B.2): f′(x) = lim_{h→0} [f(x+h) − f(x)]/h. For f(x)=x^2: 1. f(x+h) = (x+h)^2 = x^2 + 2xh + h^2. 2. Difference quotient: [(x^2 + 2xh + h^2) − x^2]/h = (2xh + h^2)/h = 2x + h. 3. Take the limit as h → 0: f′(x) = lim_{h→0} (2x + h) = 2x. So the derivative function is f′(x)=2x (also written dy/dx or y′). That value at a specific x gives the slope of the tangent line there (CHA-2.C.1). Example: at x=3, slope = 2(3)=6, so tangent line is y − 9 = 6(x − 3). If you want a quick review of this topic, check the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and try practice problems (https://library.fiveable.me/practice/ap-calculus).

I'm confused about when to use the limit definition vs derivative rules - which one should I use?

Short answer: use the limit definition when the problem specifically asks for the derivative as a limit or you need to show differentiability from first principles (CHA-2.B). Use derivative rules (power, constant multiple, sum/difference, product, quotient, chain) for speed and efficiency on routine analytic problems and most AP exam items. When to pick which: - Limit definition (f(x+h)-f(x))/h as h→0—use this to - represent f′(x) as a limit (CED CHA-2.B.2), - prove a derivative at a tricky point or show non-differentiability, - find the slope of a tangent when no rule applies. - Derivative rules—use these whenever f is built from standard pieces (polynomials, trig, exp, logs, compositions). The AP exam expects you to apply these rules quickly in multiple-choice and free-response. Tip: you should be able to do a short limit proof for at least one function (CED LO CHA-2.B) but rely on rules for most work. Review Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and drill problems at https://library.fiveable.me/practice/ap-calculus to build both skills.

Can someone walk me through finding a tangent line equation from start to finish?

Step-by-step you can always follow this 3-part recipe. 1) Find the slope (the derivative at the point). If the point is (a, f(a)), compute m = f′(a). Use the limit definition m = lim_{h→0} [f(a+h)-f(a)]/h if you must (CHA-2.B), or use derivative rules if you can. 2) Use point-slope form of the line: y − f(a) = m(x − a). (Remember f′(a) is the slope of the tangent—CHA-2.C.1.) 3) Simplify to your preferred form. Quick example: f(x)=x^2 at a=3. f′(x)=2x so m=f′(3)=6. Point (3,9). Tangent: y−9=6(x−3) → y=6x−9. For AP: show the derivative step (limit or rule) and units/interpretation if context appears (they ask for interpretation on the exam). For extra practice and examples see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and thousands of practice problems (https://library.fiveable.me/practice/ap-calculus).

What's the connection between slopes of secant lines and the derivative?

A secant line through (x, f(x)) and (x+h, f(x+h)) has slope [f(x+h) − f(x)]/h—that’s the difference quotient (average rate of change). The derivative at x, f′(x), is the limit of those secant slopes as h → 0: f′(x) = lim_{h→0} [f(x+h) − f(x)]/h (CHA-2.B.2). Geometrically, as the second point moves closer (h gets smaller), the secant line “rotates” and approaches the tangent line; the tangent’s slope is the instantaneous rate of change (CHA-2.C.1). If that limit exists, the function is differentiable at x; if secant slopes don’t settle to one value, no derivative there. Use f′(x) as the slope in the tangent-line equation y − f(x) = f′(x)(x − x0). For more AP-aligned practice and examples on limits of difference quotients, see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and try problems at (https://library.fiveable.me/practice/ap-calculus).

How do I know if I should write my answer as f'(x) or dy/dx on the AP exam?

Short answer: both are correct—use whichever fits the context and makes your reasoning clearest. How to choose: - If the problem gives a named function f(x) and asks for “the derivative as a function,” write f′(x). (CED CHA-2.B.3) - If you need the derivative at a specific x, either f′(3) or (dy/dx)|_{x=3} is fine; be consistent. - Use dy/dx (Leibniz) when doing implicit differentiation, related-rates, or when you want to emphasize the variable with respect to which you’re differentiating (e.g., dy/dt, dx/dt). Leibniz helps you track dependent/independent variables in rates problems. - Use f′(x) or y′ for concise algebraic work or when the function is named y = f(x). On the AP exam, clear notation counts (Practice 4: Communication and Notation), so pick the form that makes your steps easiest to follow and switch only when it helps (and indicate what you mean). For a quick refresher, see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and try practice problems (https://library.fiveable.me/practice/ap-calculus).

Why do I keep getting different answers when I try to evaluate the limit in the derivative definition?

If you’re getting different answers, one of these is almost always the cause: - Algebra mistakes when you simplify (expand, factor, cancel). The difference quotient needs correct algebra before you take the limit—don’t plug h = 0 too early. - Cancelling h incorrectly. For rational or root expressions you usually factor or multiply by a conjugate so an h cancels; if it doesn’t cancel the limit might not exist. - Using the wrong limit form. f'(x) = lim(h→0)[f(x+h)-f(x)]/h versus the point form lim(x→a)[f(x)-f(a)]/(x-a). They’re equivalent, but mixing steps can cause errors. - One-sided limits differ (corner, cusp, vertical tangent) → derivative doesn’t exist, so different one-sided approaches give different values. - Trig or exponential functions need identities (e.g., sin addition formula)—missing that gives wrong limits. Checklist: do algebra symbolically, cancel h, then take h→0; verify both one-sided limits at a point; if you still disagree, plug into a graphing calculator or compute the derivative with rules to compare. For worked examples and practice, see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and Unit 2 overview (https://library.fiveable.me/ap-calculus/unit-2).

What does it mean to represent a derivative graphically vs numerically vs analytically?

Graphically: f′(x) is the slope of the tangent line at each x. On a graph of f, you estimate slope by drawing tangents or by seeing where secant slopes approach a limit; on a graph of f′ you read sign (increasing/decreasing), zeros (critical points), and concavity (f′ increasing/decreasing). This matches CHA-2.C.1 and CHA-2.B.4. Numerically: you use tables or difference quotients. Approximate f′(a) with (f(a+h)−f(a))/h for small h or use symmetric/forward differences from data. That’s how the exam asks you to “estimate” derivatives from tables (Topic 2.3 connection). Analytically: you use the limit definition f′(x)=lim_{h→0}(f(x+h)−f(x))/h or derivative rules to find an exact formula f′(x) (CHA-2.B.2, CHA-2.B.3). From that you get the tangent-line equation and exact rates. On the AP exam you’ll be asked to connect these (interpret graphs/tables, compute limits, write tangent lines). For a focused review see the Topic 2.2 study guide (https://library.fiveable.me/ap-calculus/unit-2/defining-derivative-function-using-derivative-notation/study-guide/j9KaEWbB5OECijykhxCD) and try practice problems (https://library.fiveable.me/practice/ap-calculus).