2.2 Defining the Derivative of a Function and Using Derivative Notation

The derivative of a function is itself a function, defined as the limit of the difference quotient: $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$. At any point, the derivative gives the slope of the tangent line, and you can write it three common ways: $f'(x)$, $y'$, or $\frac{dy}{dx}$. For AP Calculus, recognize the same derivative idea across notation, graphs, tables, and words.

Why This Matters for the AP Calculus Exam

Unit 2 carries a noticeable weight on the AP Calculus exam (10-12% for AB, 4-7% for BC), and the limit definition of the derivative is the foundation for everything that follows. On both multiple-choice and free-response questions, you may need to recognize an expression as a derivative definition, build a difference quotient from a table or graph, switch between derivative notations, or find a tangent line equation. Later units lean on this: rates of change in context (Unit 4) and graph analysis (Unit 5) all start here.

Key Takeaways

• The derivative function is defined as $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$, provided this limit exists.
• The derivative at a point equals the slope of the tangent line at that point.
• $f'(x)$, $y'$, and $\frac{dy}{dx}$ all mean the same thing for $y = f(x)$.
• A derivative can be shown graphically, numerically, analytically, and verbally.
• To find a tangent line, get the slope from the derivative, then use point-slope form $y - y_1 = m(x - x_1)$.
• When simplifying a difference quotient, expand and combine carefully before canceling the $h$.

Definition of the Derivative

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

But how do you find the derivative for the whole curve instead of just one point? Calculating the instantaneous rate of change at every single point and then graphing would be far too tedious.

Instead, you keep the input as a variable rather than plugging in a specific number. This gives the limit definition of the derivative:

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

This produces a new function whose value at any $x$ is the slope of $f$ at that $x$, as long as the limit exists. You will learn rules and shortcuts in the next few lessons, but for now you use this definition directly.

Graph of a tangent line, created with Desmos

Here the slope of the tangent line is $f'(1)$, the derivative of $f(x)$ when $x = 1$.

Derivative Notation

There are several ways to write a derivative in calculus.

If the original function is $y = f(x)$, then the derivative can be written as $y'$, $f'(x)$, or $\frac{dy}{dx}$. These all describe the rate of change of the function as the input changes.

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

They are all valid and mean the same thing. Being comfortable switching between them helps, since exam questions can use any of these without warning.

A derivative can also appear in four representations: graphical (the slope at each point), numerical (values from a table), analytical (an algebraic formula), and verbal (a description in words). Recognizing the same idea across these forms is a skill worth building.

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Worked Examples

1) Using the Definition of a Derivative

Given $y = 3x^2 + 4x$, calculate $y'$.

Plug everything into the limit. Since $y = f(x)$,

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

Expand the numerator and distribute.

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

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

Combine like terms.

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

Since $h \neq 0$ inside the limit, divide each term by $h$.

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

As $h$ approaches $0$, the middle term approaches $0$. So:

$y' = 6x + 4$

2) Tangent Line to a Curve

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

First use the definition of the derivative.

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

Combine the two fractions in the numerator over a common denominator.

In the numerator you get $\frac{x - (x + h)}{x(x + h)}$. So:

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

Multiply the numerator by the reciprocal of the denominator.

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

Expand the numerator.

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

Cancel the $h$ from numerator and denominator.

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

As $h$ approaches $0$, the $xh$ term goes to $0$, so:

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

Now write the tangent line equation. A line in point-slope form is $y-y_1=m(x-x_1)$, where $m$ is the slope, which is the derivative at the point $(x_1,y_1)$.

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

Now you have everything for the tangent line:

$y-1=-1(x-1)$

Check that this line is tangent to $f(x)= \frac{1}{x}$ at $(1,1)$.

Graph of tangent line, created with Desmos

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How to Use This on the AP Calculus Exam

Free Response

When a question gives a table and asks you to estimate a derivative, show the difference quotient structure, not just a number. For example, write $C'(3.5) \approx \frac{C(4)-C(3)}{4-3}$ before reporting the value. A correct number without the visible setup can lose credit, so the structure is important for clear exam work.

Tangent Lines

For tangent line problems: find the slope with the derivative, evaluate it at the given point, then plug into point-slope form $y - y_1 = m(x - x_1)$. If you need a $y$-coordinate and only have $x$, get it from the original function $f$, not from $f'$.

Common Trap

When you simplify a difference quotient, keep $h$ in the denominator until you have factored it out of every term in the numerator. Canceling too early or losing a parenthesis is a frequent source of errors.

Common Misconceptions

• The derivative is just a number. The derivative $f'(x)$ is a function. Its value at a specific input is the slope at that point, but the derivative itself describes the slope across the whole domain.
• The graph of $f'$ is the same as the graph of $f$. They are different. The graph of $f'$ shows the slope of $f$ at each point, so reading $f'$ as if it were $f$ leads to wrong conclusions.
• You can cancel $h$ before simplifying. You can only divide out $h$ after it is a factor of every term in the numerator. Dividing too early gives a wrong limit.
• $\frac{dy}{dx}$ means something different from $f'(x)$. For $y = f(x)$, the notations $\frac{dy}{dx}$, $f'(x)$, and $y'$ all mean the same derivative.
• The tangent line slope comes from the original function value. The slope is the derivative at that point, not the $y$-value. Use $f$ for the point and $f'$ for the slope.

Related AP Calculus Guides

• Unit 2 Overview: Differentiation
• 2.1 Defining Average and Instantaneous Rates of Change at a Point
• 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions
• 2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist
• 2.3 Estimating Derivatives of a Function at a Point
• 2.7 Derivatives of cos x, sinx, e^x, and ln x