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What is the formal definition of the derivative, and how do you write it in the different standard notations?

Topic 2.2 Defining the Derivative of a Function and Using Derivative Notation: state the limit definition of the derivative, compute derivatives from the definition, and use standard derivative notation.

A focused answer to AP Calculus AB Topic 2.2, giving the two limit definitions of the derivative, the standard notations, and how to differentiate from first principles, with full worked examples.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. The limit definition
  3. What the derivative means
  4. Notation
  5. Recognizing a derivative hidden inside a limit
  6. Why first principles still matter

What this topic is asking

The College Board (Topic 2.2) wants you to state the limit definition of the derivative, differentiate simple functions from the definition (first principles), and write the answer in any of the standard notations. This is the formal core of differentiation that every later rule abbreviates.

The limit definition

Both forms are difference quotients in the limit. The first (hh-form) is the workhorse for finding the derivative as a function; the second (x→ax \to a form) is convenient when you want the derivative at one specific point.

What the derivative means

Notation

The same object is written several ways, and the AP exam uses all of them:

  • fβ€²(x)f'(x) - Lagrange (prime) notation.
  • dydx\frac{dy}{dx} or dfdx\frac{df}{dx} - Leibniz notation, useful for showing the variable.
  • ddx[f(x)]\frac{d}{dx}[f(x)] - the differentiation operator applied to ff.
  • yβ€²y' - shorthand when y=f(x)y = f(x).

To denote the value at a point, write fβ€²(a)f'(a) or dydx∣x=a\left.\frac{dy}{dx}\right|_{x=a}.

Recognizing a derivative hidden inside a limit

A favorite AP question gives you a limit and asks you to identify it as a derivative. The pattern to spot is exactly the definition: a difference quotient with hβ†’0h \to 0 (or xβ†’ax \to a). For instance, lim⁑hβ†’0(2+h)5βˆ’25h\lim_{h \to 0} \frac{(2 + h)^5 - 2^5}{h} is precisely fβ€²(2)f'(2) for f(x)=x5f(x) = x^5, so its value is 5(2)4=805(2)^4 = 80 without expanding anything. Reading the limit backwards - matching the base function and the point - turns an intimidating limit into a one-line derivative evaluation. Watch for the point baked into the expression (here the 22) and the function suggested by the form (here a fifth power). This skill connects Unit 1 limits directly to Unit 2 derivatives and rewards students who see the definition as a template rather than a one-off calculation.

Why first principles still matter

Later topics give shortcut rules (power, product, quotient) that bypass the limit, but the AP exam still asks you to recognize and apply the definition - for example to identify a limit as a derivative, or to differentiate a function for which you only know the definition. Understanding that every rule is just a shorthand for this limit keeps the rules from feeling arbitrary. The two equivalent forms each have a natural use: the h→0h \to 0 form gives the derivative as a whole new function of xx, which is what you want when you will evaluate at several points or analyze behavior, while the x→ax \to a form is convenient when a single value f′(a)f'(a) is all that is needed. Being fluent in moving between them, and in reading a given limit as one of them, is what the definition-of-the-derivative questions are really testing.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (style)1 marksSection I (multiple choice, no calculator). Which limit equals fβ€²(a)f'(a)? (A) lim⁑hβ†’0f(a)βˆ’f(a+h)h\lim_{h \to 0} \frac{f(a) - f(a+h)}{h} (B) lim⁑hβ†’0f(a+h)βˆ’f(a)h\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} (C) lim⁑xβ†’0f(x)βˆ’f(a)xβˆ’a\lim_{x \to 0} \frac{f(x) - f(a)}{x - a} (D) f(a+h)βˆ’f(a)h\frac{f(a+h) - f(a)}{h}
Show worked answer β†’

The correct answer is (B).

The limit definition of the derivative at aa is fβ€²(a)=lim⁑hβ†’0f(a+h)βˆ’f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. (A) has the numerator reversed (it would give βˆ’fβ€²(a)-f'(a)). (C) uses the wrong limit point (xβ†’ax \to a, not xβ†’0x \to 0). (D) is the difference quotient without the limit, so it is just an average rate, not the derivative.

AP 2023 (style)3 marksSection II (free response, no calculator). Let f(x)=x2βˆ’3xf(x) = x^2 - 3x. (a) Write the limit definition of fβ€²(x)f'(x). (b) Compute fβ€²(x)f'(x) from the definition. (c) Evaluate fβ€²(2)f'(2) and interpret it as a slope.
Show worked answer β†’

A 3-point first-principles question.

(a) (1 point) fβ€²(x)=lim⁑hβ†’0f(x+h)βˆ’f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
(b) (1 point) f(x+h)=(x+h)2βˆ’3(x+h)=x2+2xh+h2βˆ’3xβˆ’3hf(x+h) = (x+h)^2 - 3(x+h) = x^2 + 2xh + h^2 - 3x - 3h, so f(x+h)βˆ’f(x)=2xh+h2βˆ’3hf(x+h) - f(x) = 2xh + h^2 - 3h. Then 2xh+h2βˆ’3hh=2x+hβˆ’3β†’2xβˆ’3\frac{2xh + h^2 - 3h}{h} = 2x + h - 3 \to 2x - 3. So fβ€²(x)=2xβˆ’3f'(x) = 2x - 3.
(c) (1 point) fβ€²(2)=2(2)βˆ’3=1f'(2) = 2(2) - 3 = 1; the tangent line to y=f(x)y = f(x) at x=2x = 2 has slope 11.

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