Defining the derivative and using derivative notation — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: This chapter covers the two limit-based definitions of the derivative, Leibniz and prime derivative notation, alternate forms for the derivative at a point, and recognizing derivatives from given limit expressions, aligned to the AP Calculus AB CED.

You should already know: How to evaluate one-sided and two-sided limits algebraically and graphically. The formula for the slope of a secant line between two points on a function. Basic function notation and algebraic simplification techniques.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Defining the derivative and using derivative notation?

The derivative is the core tool of calculus, measuring the instantaneous rate of change of a function at a given point. Per the AP Calculus AB CED, this subtopic falls within Unit 2 (Differentiation: Definition and Fundamental Properties), which accounts for 10–12% of the total AP exam score; this specific topic typically contributes 1–2 multiple-choice (MCQ) questions and occasionally a part of an early free-response (FRQ) question, so it appears in both sections of the exam. It connects the precalculus concept of slope between two points to the calculus idea of instantaneous change at a single point, meaning all subsequent differentiation rules and applications build directly on this definition. In this chapter, we cover two equivalent limit forms for the derivative (the derivative at a point, and the derivative as a function), all standard notations used on the AP exam, and how to match notation and limit expressions to the correct derivative.

2. The Derivative at a Point

The most fundamental definition of a derivative describes the slope of the tangent line to a function $f(x)$ at a specific point $x=a$. To get this slope, we start with the slope of a secant line between $(a,f(a))$ and $(a+h,f(a+h))$, where $h$ is the step size between the two points. As $h$ approaches 0, the two points get closer together, and the secant slope approaches the tangent slope (the derivative). The formal definition is: $$f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}$$ An equivalent alternate form, often used for recognition problems rather than calculation, replaces the step size $h$ with a variable $x$ approaching $a$: $$f^{\prime }(a)=\underset{x\to a}{\lim }\frac{f(x)-f(a)}{x-a}$$ The derivative only exists at $x=a$ if this two-sided limit exists. If the limit exists, we say $f$ is differentiable at $x=a$. The value $f^{\prime }(a)$ is a constant equal to the instantaneous rate of change of $f(x)$ at $x=a$, as well as the slope of the tangent line at that point.

Worked Example

Use the limit definition to calculate $f^{\prime }(2)$ for $f(x)=3x^2-5x$.

1. Substitute into the definition with $a=2$: $f^{\prime }(2)={\lim }_{h\to 0}\frac{f(2+h)-f(2)}{h}$
2. Expand $f(2+h)$: $f(2+h)=3(2+h{)}^2-5(2+h)=12+12h+3h^2-10-5h=2+7h+3h^2$
3. Calculate $f(2)=3(2{)}^2-5(2)=12-10=2$, so the numerator simplifies to: $f(2+h)-f(2)=7h+3h^2=h(7+3h)$
4. Cancel $h$ (valid for $h\ne 0$, which holds when taking the limit as $h\to 0$): $\frac{h(7+3h)}{h}=7+3h$
5. Evaluate the limit: ${\lim }_{h\to 0}(7+3h)=7$, so $f^{\prime }(2)=7$.

Exam tip: If an AP question explicitly asks you to use the limit definition to find a derivative, you will earn zero credit for only using a shortcut differentiation rule; you must show the full limit calculation to earn points.

3. The Derivative as a Function and Derivative Notation

Once we can calculate the derivative at any single point $x=a$, we can treat the derivative as a function itself, where the input is any $x$ where the derivative exists, and the output is the derivative at that $x$. The formal definition of the derivative as a function is: $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$ The set of all $x$ where this limit exists is the domain of $f^{\prime }(x)$. Two standard notations for derivatives are used interchangeably on the AP exam:

1. Prime notation: For a function $y=f(x)$, the derivative is written $f^{\prime }(x)$ (read "f prime of x") or $y^{\prime }$. For the derivative at $x=a$, this becomes $f^{\prime }(a)$ or $y^{\prime }(a)$.
2. Leibniz notation: For a function $y=f(x)$, the derivative is written $\frac{dy}{dx}$ (read "dee y dee x") or $\frac{d}{dx}[f(x)]$, where $\frac{d}{dx}$ denotes the operation of taking the derivative. For the derivative at $x=a$, this is written $\frac{dy}{dx}{|}_{x=a}$, with the evaluation bar indicating the derivative is evaluated at $x=a$.

Worked Example

Use the limit definition to find $f^{\prime }(x)$ for $f(x)=\sqrt{2x}$.

1. Substitute into the general limit definition: $f^{\prime }(x)={\lim }_{h\to 0}\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h}$
2. Rationalize the numerator by multiplying numerator and denominator by the conjugate $\sqrt{2x+2h}+\sqrt{2x}$: $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{(\sqrt{2x+2h}-\sqrt{2x})(\sqrt{2x+2h}+\sqrt{2x})}{h(\sqrt{2x+2h}+\sqrt{2x})}$$
3. Simplify the numerator (difference of squares): $(2x+2h)-2x=2h$, so we get: $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{2h}{h(\sqrt{2x+2h}+\sqrt{2x})}$$
4. Cancel $h$ (valid for $h\ne 0$) and evaluate the limit by substituting $h=0$: $$f^{\prime }(x)=\frac{2}{2\sqrt{2x}}=\frac{1}{\sqrt{2x}}$$

Exam tip: When writing the derivative evaluated at a point in Leibniz notation, always place the evaluation bar after the derivative expression; misplacing the bar will be marked incorrect on FRQs.

4. Recognizing Derivatives from Limit Expressions

A common AP exam question gives you an unsimplified limit of a difference quotient and asks you to either identify what derivative it represents or evaluate the limit using your knowledge of derivatives, instead of calculating the limit from scratch. This tests whether you understand the structure of the derivative definition, not just how to compute with it. To solve these problems, match the given limit to the structure of one of the two derivative definitions:

• For a limit of the form ${\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$, the limit equals $f^{\prime }(a)$
• For a limit of the form ${\lim }_{x\to a}\frac{f(x)-f(a)}{x-a}$, the limit also equals $f^{\prime }(a)$ Once you identify $f(x)$ and $a$, you can evaluate the derivative using differentiation rules to get the value of the limit, without computing the limit by hand.

Worked Example

Evaluate ${\lim }_{h\to 0}\frac{(1+h{)}^3-1}{h}$ by recognizing it as a derivative.

1. Match the given limit to the form ${\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$. The term $(1+h{)}^3$ matches $f(a+h)$, so $a=1$ and $f(x)=x^3$. The constant term $1$ equals $f(1)=1^3=1$, so the match is correct.
2. The limit is therefore equal to $f^{\prime }(1)$ for $f(x)=x^3$.
3. Using the power rule for differentiation, $f^{\prime }(x)=3x^2$, so evaluate at $x=1$: $f^{\prime }(1)=3(1{)}^2=3$.
4. Confirm by direct calculation: ${\lim }_{h\to 0}\frac{1+3h+3h^2+h^3-1}{h}={\lim }_{h\to 0}(3+3h+h^2)=3$, which matches.

Exam tip: Always check that $f(a)$ matches the constant term in the numerator; common MCQ distractors use the wrong constant term to test if you match all parts of the definition correctly.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Substituting $h=0$ before simplifying the difference quotient, leading to an undefined $0/0$ result. Why: Students confuse the order of operations for limits, thinking substitution is always the first step. Correct move: Always simplify the difference quotient by factoring, combining fractions, or rationalizing to cancel $h$ first, then substitute $h=0$ to evaluate the limit.
• Wrong move: Identifying ${\lim }_{x\to 2}\frac{f(x)-f(1)}{x-2}$ as $f^{\prime }(1)$. Why: Students match the constant in the numerator to the point, ignoring the denominator. Correct move: For the alternate derivative form, the denominator is $x-a$, so the point is $a$, regardless of where the constant appears in the numerator; this specific limit is not a valid derivative of $f$.
• Wrong move: Leaving $x$ as a variable when asked for $f^{\prime }(a)$, or substituting $a$ early when asked for the general derivative function $f^{\prime }(x)$. Why: Students mix up the two types of derivative problems after practicing both. Correct move: Always re-read the question to confirm: if it asks for the derivative at a point, your final answer is a constant; if it asks for the derivative function, your final answer is a function of $x$.
• Wrong move: Claiming $f^{\prime }(0)$ exists just because ${\lim }_{x\to 0}f(x)$ exists. Why: Students confuse the existence of the limit of the original function with the existence of the derivative. Correct move: The derivative only exists if the two-sided limit of the difference quotient exists, regardless of whether the limit of the original function exists.
• Wrong move: Writing $\frac{dy(3)}{dx}$ to denote the derivative of $y=f(x)$ at $x=3$. Why: Students misapply fraction notation to derivative evaluation. Correct move: Always use an evaluation bar after the derivative: $\frac{dy}{dx}{|}_{x=3}$.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following correctly denotes the derivative of $f(x)=\ln x$ evaluated at $x=3$? A) $\frac{d}{dx}\ln 3$ B) $\frac{d(\ln 3)}{dx}$ C) $\frac{d}{dx}(\ln x){|}_{x=3}$ D) $\frac{dy}{dx}(\ln 3)$

Worked Solution: We need the derivative of the function $\ln x$, evaluated at $x=3$. Options A, B, and D all incorrectly treat $\ln 3$, a constant, as the function to differentiate. The derivative of a constant is zero, so these do not give the derivative of $\ln x$ at $x=3$. Only Option C correctly denotes taking the derivative of $\ln x$, then evaluating the resulting derivative at $x=3$. Correct answer: C

Question 2 (Free Response)

Let $f(x)=\frac{2}{x+1}$. (a) Use the limit definition of the derivative to find $f^{\prime }(x)$. Show all your work. (b) Write $f^{\prime }(1)$ in two different standard notations. (c) Explain what $f^{\prime }(1)$ means in terms of the tangent line to $f(x)$.

Worked Solution: (a) Start with the general limit definition: $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{\lim }\frac{\frac{2}{x+h+1}-\frac{2}{x+1}}{h}$$ Combine the fractions in the numerator: $$=\underset{h\to 0}{\lim }\frac{2(x+1)-2(x+h+1)}{h(x+h+1)(x+1)}=\underset{h\to 0}{\lim }\frac{-2h}{h(x+h+1)(x+1)}$$ Cancel $h$ (valid for $h\ne 0$) and evaluate the limit: $$=\underset{h\to 0}{\lim }\frac{-2}{(x+h+1)(x+1)}=\frac{-2}{(x+1{)}^2}$$ Thus $f^{\prime }(x)=\frac{-2}{(x+1{)}^2}$.

(b) Two valid notations are: $f^{\prime }(1)$ (prime notation) and $\frac{d}{dx}\left(\frac{2}{x+1}\right){|}_{x=1}$ (Leibniz notation).

(c) $f^{\prime }(1)=\frac{-2}{(1+1{)}^2}=-\frac{1}{2}$. This means the slope of the tangent line to $y=f(x)$ at the point $(1,f(1))=(1,1)$ is $-\frac{1}{2}$.

Question 3 (Application / Real-World Style)

The height of a projectile $t$ seconds after being launched is given by $h(t)=-16t^2+48t+10$, where $h(t)$ is measured in feet. Use the limit definition of the derivative at a point to find $h^{\prime }(2)$, and interpret your result in context.

Worked Solution: Use the definition $h^{\prime }(a)={\lim }_{h\to 0}\frac{h(a+h)-h(a)}{h}$ with $a=2$:

1. Calculate $h(2)=-16(2{)}^2+48(2)+10=-64+96+10=42$
2. Expand $h(2+h)=-16(2+h{)}^2+48(2+h)+10=-16(4+4h+h^2)+96+48h+10=-64-64h-16h^2+106+48h=42-16h-16h^2$
3. Simplify the difference quotient: $\frac{h(2+h)-h(2)}{h}=\frac{(42-16h-16h^2)-42}{h}=-16-16h$
4. Evaluate the limit: ${\lim }_{h\to 0}(-16-16h)=-16$, so $h^{\prime }(2)=-16$ feet per second.

Interpretation: At 2 seconds after launch, the height of the projectile is decreasing at a rate of 16 feet per second.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for all of differentiation, one of the two core pillars of AP Calculus AB. Immediately after this, you will learn the shortcut differentiation rules that let you compute derivatives far faster than using the limit definition every time. Without mastering the limit definition of the derivative and the meaning of notation, you will not be able to recognize when a question asks for a derivative, or understand where the shortcut rules come from. This topic also feeds into all later applications of derivatives: tangent lines, related rates, optimization, and the Fundamental Theorem of Calculus, all of which depend on the derivative being a limit that measures instantaneous rate of change.

Basic differentiation rules: power, constant, sum, constant multiple Product and quotient rules for differentiation Differentiability and continuity Derivatives of composite functions: the chain rule