AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/2/2-2)
Unit 2 · Topic 2.2
2.2 Defining the Derivative of a Function and Using Derivative Notation
The derivative of f at x = a is the limit of the difference quotient: the exact instantaneous rate of change and the slope of the tangent line there. You need both limit forms of the definition, the standard notation, and the skill of spotting a derivative hidden inside a limit.
Key terms
- derivative
- limit definition of the derivative
- f′(x)
- dy/dx
- slope of the tangent line
The definition, two ways
The derivative of f at x = a is f′(a) = lim (h→0) (f(a + h) − f(a)) / h, provided the limit exists. An equivalent form uses a second point x approaching a: f′(a) = lim (x→a) (f(x) − f(a)) / (x − a).
If the limit exists, f is differentiable at a. Geometrically, f′(a) is the slope of the tangent line to the graph at (a, f(a)).
The derivative as a function
Let a vary and you get a new function, f′(x) = lim (h→0) (f(x + h) − f(x)) / h. Its input is an x-value, and its output is the slope of f at that x. The domain of f′ is every x where that limit exists.
For f(x) = x², the difference quotient simplifies to ((x + h)² − x²)/h = (2xh + h²)/h = 2x + h, which approaches 2x. So f′(x) = 2x: the slope at x = 3 is 6, and at x = −1 it is −2.
Notation you need to read and write
All of these name the same idea. Leibniz notation is especially handy in context: dV/dt is the rate the volume V changes with respect to time t.
| Notation | Read as | Notes |
|---|---|---|
| f′(x) | f prime of x | Lagrange (prime) notation |
| y′ | y prime | When y = f(x) |
| dy/dx | dee y dee x | Leibniz notation: rate of change of y with respect to x |
| d/dx [f(x)] | the derivative of f(x) | An instruction to differentiate |
| dy/dx evaluated at x = a | the derivative at a | Same as f′(a) |
Recognizing a derivative in disguise
Exam questions often hand you a limit that is secretly a derivative. Match it to one of the two forms. For example, lim (h→0) (√(9 + h) − 3) / h has the shape (f(a + h) − f(a)) / h with f(x) = √x and a = 9. So it equals f′(9).
Likewise, lim (x→π) (cos x + 1) / (x − π) has the shape (f(x) − f(a)) / (x − a) with f(x) = cos x and a = π, because cos π = −1. Once you know the derivative rules from the coming topics, these limits are quick.
The tangent line
The tangent line at x = a passes through (a, f(a)) with slope f′(a). In point-slope form: y − f(a) = f′(a)(x − a). You'll use this equation all year, especially for approximations in 4.6.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Using the definition to find f′(2)
Use the limit definition to find f′(2) for f(x) = x² − 3x.
Show the solutionHide the solution
- Step 1: f(2) = 4 − 6 = −2 and f(2 + h) = (2 + h)² − 3(2 + h) = 4 + 4h + h² − 6 − 3h = h² + h − 2.
- Step 2: Difference quotient: (f(2 + h) − f(2)) / h = (h² + h − 2 − (−2)) / h = (h² + h) / h = h + 1, for h ≠ 0.
- Step 3: Take the limit: lim (h→0) (h + 1) = 1.
Answer: f′(2) = 1
- Example 2
A limit that is a derivative
Find lim (h→0) (1/(2 + h) − 1/2) / h by recognizing it as a derivative.
Show the solutionHide the solution
- Step 1: Match the form (f(a + h) − f(a)) / h: here f(x) = 1/x and a = 2.
- Step 2: So the limit equals f′(2). To compute it from the limit directly, combine fractions: (2 − (2 + h)) / (2(2 + h)) = −h / (2(2 + h)).
- Step 3: Divide by h: −1 / (2(2 + h)), which approaches −1/4 as h→0.
- Step 4: Check with the power rule (2.5): f(x) = x⁻¹ gives f′(x) = −x⁻², and f′(2) = −1/4.
Answer: The limit is f′(2) for f(x) = 1/x, which equals −1/4.
- Example 3
Trap: matching the wrong function
Find lim (x→π) (cos x + 1) / (x − π).
Show the solutionHide the solution
- Step 1: A quick guess might be f(x) = cos x + 1, but think of it as (f(x) − f(π)) / (x − π) with f(x) = cos x. Since f(π) = cos π = −1, the top f(x) − f(π) = cos x − (−1) = cos x + 1. It matches.
- Step 2: So the limit is f′(π). The derivative of cos x is −sin x (2.7).
- Step 3: f′(π) = −sin π = 0.
Answer: 0
Common mistakes
- Expanding f(a + h) incorrectly, like writing (2 + h)² = 4 + h². Expand fully: 4 + 4h + h².
- Dropping f(a) or the limit symbol partway through. Keep lim (h→0) written until you substitute.
- Treating f′(a) as the y-value of the tangent line. It's the slope; the point on the line is (a, f(a)).
On the exam
- Multiple-choice questions often show a limit like lim (h→0) (sin(π/3 + h) − sin(π/3))/h and ask for its value. Identify f and a, then use derivative rules.
- Free-response questions ask for tangent lines constantly. Show the point, the slope and the equation.
Connected topics
Videos
Check yourself
4 questions on 2.2 Defining the Derivative of a Function and Using Derivative Notation. Pick an answer to see if you got it, and why.
What is lim (h→0) ((2 + h)⁴ − 16)/h ?
The limit lim (h→0) (e^(2 + h) − e²)/h is equal to f′(a). Which of the following could be the function f and the number a? I. f(x) = eˣ and a = 2 II. f(x) = e^(x + 2) and a = 0 III. f(x) = e²ˣ and a = 1
The function f is differentiable, and f′(2) = 5. What is lim (h→0) (f(2 + 3h) − f(2))/h ?
The function f is differentiable, and f(3) = 5. The line tangent to the graph of f at x = 3 passes through the point (1, 1). What is the value of f′(3)?
0 of 4 answered