Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). $$ s(x)=2 x+1 $$

Step-by-Step Solution

Answer

The derivative of \( s(x) = 2x + 1 \) is \( 2 \).

Step 1: Substitute the Function into the Derivative Formula

Since the function is given as \(s(x) = 2x + 1\), we start by substituting this into the derivative formula: \[ f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Becomes: \[ s^{\prime}(x) = \lim_{h \to 0} \frac{(2(x+h) + 1) - (2x + 1)}{h} \]

Step 2: Simplify the Expression in the Numerator

Expand and simplify the numerator: \[ (2(x+h) + 1) - (2x + 1) = 2x + 2h + 1 - 2x - 1 \] \[ = 2h \]

Step 3: Divide by h and Take the Limit

Substitute the simplified numerator back into the limit expression: \[ s^{\prime}(x) = \lim_{h \to 0} \frac{2h}{h} \] Cancel out \( h \): \[ = \lim_{h \to 0} 2 \] Since there are no more \( h \) terms remaining, the derivative is simply \( 2 \).

Key Concepts

Limit Definition of Derivative

To find the derivative of a function, calculus uses a fundamental concept known as the limit definition of derivative. This idea focuses on measuring the instantaneous rate of change of the function at a certain point.
The formula used is:

• \( f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

Here, the derivative \( f^{\prime}(x) \) is found by considering what happens when \( h \), a small increment, approaches zero.
As \( h \) gets infinitesimally small, the quotient \( \frac{f(x+h) - f(x)}{h} \) provides the slope of the tangent line to the curve at the point \( x \).
Understanding this concept is crucial because it lays the foundation for differentiation and underpins many principles in calculus.

Differentiation

Differentiation is the process of finding the derivative of a function.
It's a technique in calculus that helps us find how a function changes, its rate of change, and the slope of its graph.
Differentiation transforms complex functions into simpler expressions that are easier to analyze.
When applying the differentiation process to a linear function like \( s(x) = 2x + 1 \), we use the limit definition as a step-by-step method:

• First, substitute the function into the derivative formula.
• Simplify the expression in the numerator of the fraction.
• Finally, take the limit as \( h \) approaches zero.

Differentiation is a cornerstone of calculus, enabling mathematicians and scientists to make precise predictions about changing systems.

Linear Functions

Linear functions are among the simplest types of functions you will encounter.
They are represented in the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
The graph of a linear function is a straight line because the rate of change remains constant.
For example, in the function \( s(x) = 2x + 1 \), the slope \( m \) is 2, and the y-intercept \( b \) is 1.
The slope \( m \) tells us how steep the line is and what we find when we derive it:
The derivative \( s^{\prime}(x) = 2 \), confirms that the slope of this linear function is constant and equal to 2.
Understanding linear functions is critical for solving calculus problems as they often serve as the backbone for more complex functions.

Calculus Problems

Solving calculus problems often involves using various principles, including differentiation, to understand and describe real-world scenarios.
Calculus enables us to model and solve complex problems involving rates of change and slopes.
While some problems may seem daunting at first, breaking them down into smaller parts often makes them more manageable.

• Start by identifying the function and what it's describing.
• Use differentiation to find the rate of change or slope.
• Simplify each step for clarity.
• Apply the results to the problem context.

By using approaches like these, calculus allows us to tackle a wide range of challenges, from physics and engineering to economics and beyond.