Defining the Derivative of a Function and Using Derivative Notation - AP Calculus BC Study Guide

The Core Idea: Defining the Derivative of a Function and Using Derivative Notation

The central concept of this topic is the formal definition of the derivative. The derivative of a function at a point represents the instantaneous rate of change of the function at that point, which is geometrically interpreted as the slope of the line tangent to the function's graph. This instantaneous rate is determined by considering the average rate of change between two points on the curve and then taking the limit as the distance between these points approaches zero.

This process transforms the slope of a secant line, $\frac{f(x+h)-f(x)}{h}$, into the slope of the tangent line by letting the interval $h$ shrink to zero. The result is a new function, called the derivative, which gives the slope of the original function at any given point. This topic establishes the fundamental limit definitions that serve as the foundation for all differentiation rules that follow.

Key Definitions

The derivative of a function can be defined using two primary, equivalent limit expressions. It is also represented using several standard notations.

The Limit Definition of the Derivative (as a function)

The derivative of a function $f$, denoted $f^{\prime }$, is a function whose value at $x$ is given by:

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

This definition is used to find a general formula for the derivative function, which can then be evaluated at any point in its domain.

The Alternative Limit Definition of the Derivative (at a point)

The derivative of a function $f$ at a specific point $x=a$, denoted $f^{\prime }(a)$, is given by:

$$f^{\prime }(a)=\underset{x\to a}{\lim }\frac{f(x)-f(a)}{x-a}$$

This definition is often used when you only need to find the derivative at a single, specific point.

Derivative Notation

The derivative of a function $y=f(x)$ can be expressed in several ways:

• Prime Notation:$f^{\prime }(x)$ or $y^{\prime }$

• Leibniz Notation:$\frac{dy}{dx}$

Understanding the Derivative as a Function

A critical concept is that the derivative of a function, $f(x)$, is itself a function. The expression $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$ takes an input $x$ and produces an output, $f^{\prime }(x)$. This output value is not the value of the original function, but rather the slope of the tangent line to the graph of $y=f(x)$ at that specific input x`. For example, if $f(x) = x^2, its derivative is the function $f^{\prime }(x)=2x$. This means:

• At $x=1$, the slope of the tangent line to $f(x)$ is $f^{\prime }(1)=2(1)=2$.

• At $x=-3$, the slope of the tangent line to $f(x)$ is $f^{\prime }(-3)=2(-3)=-6$.

The function $f^{\prime }(x)$ provides a complete map of the instantaneous rate of change for every point on the original function $f(x)$.

Core Concepts & Rules

• The derivative of a function $f$ is a new function, $f^{\prime }$, defined by the limit $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$.

• This limit represents the instantaneous rate of change of $f$ with respect to $x$.

• The derivative at a specific point $x=a$ can be found using the alternative limit definition: $f^{\prime }(a)={\lim }_{x\to a}\frac{f(x)-f(a)}{x-a}$.

• Common notations for the derivative of a function $y=f(x)$ include $f^{\prime }(x)$, $y^{\prime }$, and $\frac{dy}{dx}$. Each notation is used in different contexts but represents the same concept.

Step-by-Step Example 1: Finding the Derivative Function Using the Limit Definition

Problem: Find the derivative of the function $f(x)=3x^2-2x$ using the limit definition $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$.

Step 1: Find $f(x+h)$

Substitute $(x+h)$ for every $x$ in the original function.

$f(x+h)=3(x+h{)}^2-2(x+h)$

$f(x+h)=3(x^2+2xh+h^2)-2x-2h$

$f(x+h)=3x^2+6xh+3h^2-2x-2h$

Step 2: Set up the difference quotient $\frac{f(x+h)-f(x)}{h}$

Substitute the expressions for $f(x+h)$ and $f(x)$ into the formula. Be careful with parentheses around $f(x)$.

$$\frac{(3x^2+6xh+3h^2-2x-2h)-(3x^2-2x)}{h}$$

Step 3: Simplify the numerator

Distribute the negative sign and combine like terms. All terms that do not contain an $h$ should cancel out.

$$\frac{3x^2+6xh+3h^2-2x-2h-3x^2+2x}{h}$$

$$\frac{6xh+3h^2-2h}{h}$$

Step 4: Factor out and cancel $h$

Factor $h$ from each term in the numerator. Since $h\to 0$, $h$ is not equal to zero, so we can cancel.

$$\frac{h(6x+3h-2)}{h}$$

$$6x+3h-2$$

Step 5: Evaluate the limit

Now, take the limit as $h\to 0$.

$$f^{\prime }(x)=\underset{h\to 0}{\lim }(6x+3h-2)$$

Substitute $h=0$:

$$f^{\prime }(x)=6x+3(0)-2=6x-2$$

Solution: The derivative of $f(x)=3x^2-2x$ is $f^{\prime }(x)=6x-2$.

Step-by-Step Example 2: Finding the Derivative at a Point Using the Alternative Definition

Problem: Find the slope of the tangent line to the graph of $f(x)=\frac{1}{x+1}$ at the point $x=2$.

Step 1: Identify the components for the alternative definition

We need to find $f^{\prime }(2)$. The alternative definition is $f^{\prime }(a)={\lim }_{x\to a}\frac{f(x)-f(a)}{x-a}$.

Here, $a=2$, $f(x)=\frac{1}{x+1}$, and $f(a)=f(2)=\frac{1}{2+1}=\frac{1}{3}$.

Step 2: Set up the limit expression

Substitute the components into the formula.

$$f^{\prime }(2)=\underset{x\to 2}{\lim }\frac{\frac{1}{x+1}-\frac{1}{3}}{x-2}$$

Step 3: Simplify the complex fraction

Find a common denominator for the terms in the numerator, which is $3(x+1)$.

$$f^{\prime }(2)=\underset{x\to 2}{\lim }\frac{\frac{3}{3(x+1)}-\frac{x+1}{3(x+1)}}{x-2}$$

$$f^{\prime }(2)=\underset{x\to 2}{\lim }\frac{\frac{3-(x+1)}{3(x+1)}}{x-2}$$

$$f^{\prime }(2)=\underset{x\to 2}{\lim }\frac{\frac{2-x}{3(x+1)}}{x-2}$$

Step 4: Rewrite the expression to cancel the $(x-2)$ term

Factor a $-1$ from the numerator's numerator to reveal a $(x-2)$ term.

$$f^{\prime }(2)=\underset{x\to 2}{\lim }\frac{\frac{-(x-2)}{3(x+1)}}{x-2}$$

Now, rewrite the division as multiplication by the reciprocal.

$$f^{\prime }(2)=\underset{x\to 2}{\lim }\frac{-(x-2)}{3(x+1)}\cdot \frac{1}{x-2}$$

Cancel the $(x-2)$ terms.

$$f^{\prime }(2)=\underset{x\to 2}{\lim }\frac{-1}{3(x+1)}$$

Step 5: Evaluate the limit

Substitute $x=2$ into the simplified expression.

$$f^{\prime }(2)=\frac{-1}{3(2+1)}=\frac{-1}{3(3)}=-\frac{1}{9}$$

Solution: The slope of the tangent line at $x=2$ is $-\frac{1}{9}$.

Using Your Calculator

The definitions of the derivative covered in this topic are analytical. You are expected to compute these limits by hand using algebraic manipulation. A calculator should not be used to find the answer, but it can be an excellent tool for checking your work.

To check the result of Example 2 ($f^{\prime }(2)$ for $f(x)=\frac{1}{x+1}$):

1. Press the $[math]$ key.

2. Scroll down to 8:nDeriv( or press the number $8$.

3. The syntax is nDeriv(function, variable, value).

4. Enter the expression: nDeriv(1/(X+1), X, 2)

5. Press [ENTER]. The calculator will return a value very close to $-1/9$, such as $-0.111111111$. This confirms your analytical result.

AP Exam Quick Hit

Common Question Types

• Recognizing a Limit as a Derivative: You will be given a limit and asked to evaluate it. The key is to recognize it as one of the derivative definitions, identify the function $f(x)$ and the point $a$, and then use simpler differentiation rules (learned later) to find the value.

  • Example: Evaluate ${\lim }_{h\to 0}\frac{\cos (\frac{\pi }{2}+h)-\cos (\frac{\pi }{2})}{h}$. You should recognize this as the derivative of $f(x)=\cos (x)$ at $x=\frac{\pi }{2}$.

• Direct Calculation from the Definition: A free-response question may explicitly state: "Using the definition of the derivative, find $f^{\prime }(x)$." This forbids the use of shortcut rules and requires you to show the full limit calculation, similar to Example 1 above.

  • Example: Let $f(x)=\sqrt{x+5}$. Find $f^{\prime }(x)$ using the limit definition of the derivative.

Common Mistakes

• Algebraic Errors: The most frequent mistakes are algebraic, especially when expanding $f(x+h)$. Forgetting to distribute a negative sign across all terms of $f(x)$ is a classic error.

• Dropping the Limit Notation: Writing $\frac{f(x+h)-f(x)}{h}=\cdots =6x-2$ is mathematically incorrect. The ${\lim }_{h\to 0}$ notation must be carried through on each step of the calculation until the limit is actually evaluated (i.e., when $h=0$ is substituted).

• Incorrect Substitution: Confusing $f(x+h)$ with $f(x)+h$. For $f(x)=x^2$, $f(x+h)=(x+h{)}^2$, not $x^2+h$.

• Premature Substitution: Plugging in $h=0$ at the beginning of the problem, which results in the indeterminate form $\frac{0}{0}$. The goal of the algebraic simplification is to eliminate the $h$ in the denominator so that direct substitution becomes possible.