Problem 56 - Calculus Volume 1

Question

Use the definition of a derivative to find \(f^{\prime}(x)\). $$ f(x)=\frac{2 x}{7}+1 $$

Step-by-Step Solution

Answer

The derivative is \(\frac{2}{7}\).

Step 1: Understand the Definition of a Derivative

The definition of a derivative at a point is given by the limit: \(f^{\prime}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\). This is the foundational concept that we'll use to find the derivative.

Step 2: Calculate \(f(x+h)\)

First, substitute \(x + h\) into the function \(f(x) = \frac{2x}{7} + 1\). We get \(f(x + h) = \frac{2(x + h)}{7} + 1\). Simplifying this, we find: \(f(x + h) = \frac{2x + 2h}{7} + 1\).

Step 3: Substitute into the Derivative Definition

Substitute \(f(x + h)\) and \(f(x)\) into the definition that's \(\frac{f(x+h) - f(x)}{h}\). This yields: \(\frac{\left(\frac{2x + 2h}{7} + 1\right) - \left(\frac{2x}{7} + 1\right)}{h}\).

Step 4: Simplify the Expression

Simplify \(\frac{\left(\frac{2x + 2h}{7} + 1\right) - \left(\frac{2x}{7} + 1\right)}{h}\). The expression inside the numerator becomes \(\frac{2x + 2h}{7} + 1 - \frac{2x}{7} - 1\). Cancel terms to simplify to \(\frac{2h}{7}\).

Step 5: Evaluate the Limit

Now, simplify the expression further: \(\frac{2h}{7h}\). The \(h\) in the numerator and denominator cancels, giving \(\frac{2}{7}\). The limit as \(h \to 0\) of \(\frac{2}{7}\) is simply \(\frac{2}{7}\).

Step 6: Final Answer

Therefore, the derivative \(f^{\prime}(x)\) of the function \(\frac{2x}{7} + 1\) is \(\frac{2}{7}\).

Key Concepts

Definition of a Derivative

To understand the concept of a derivative, it's crucial to start with its definition. The derivative of a function at any point gives us the rate at which the function is changing at that point. Think of it like observing how steep a hill is at a specific location. This fundamental concept in calculus is expressed through a limit. The formal definition is:

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

Here, \(h\) represents a small increment in \(x\), and \(f(x+h) - f(x)\) denotes the change in the function's value as \(x\) increases by \(h\). The limit process as \(h\) approaches zero captures the behavior of the function's slope at ever-smaller scales, effectively providing the instantaneous rate of change.

Limit Process

The limit process is the bridge to connect how functions behave at a point. When exploring derivatives, we use limits to understand the tiny changes in the function as \(h\) approaches zero. To apply this on our function, we first assess the function at an incremented point, \(f(x+h)\).
For instance, for \(f(x) = \frac{2x}{7} + 1\), we first find \(f(x + h)\):

• Substitute \(x + h\) to get \(f(x+h) = \frac{2(x + h)}{7} + 1\).
• This simplifies to \(\frac{2x + 2h}{7} + 1\).

By evaluating the function at \(x + h\) and comparing it with its value at \(x\), we effectively construct a small segment of the function over which to measure its rate of change. As we simplify \(\frac{f(x+h) - f(x)}{h}\), we are preparing to tackle the limit as \(h\) approaches zero, ultimately revealing the function's derivative.

Simplifying Expressions

Simplification plays a pivotal role in finding derivatives. It helps to untangle complex expressions. In the context of derivatives, we often have to simplify functions before taking limits.
Consider the expression:

• \(\frac{\left(\frac{2x + 2h}{7} + 1\right) - \left(\frac{2x}{7} + 1\right)}{h}\).

This looks tricky at first, but careful subtraction and simplification make it manageable.

• The terms \(\frac{2x + 2h}{7} + 1 - \frac{2x}{7} - 1\) show that most parts cancel each other out, leaving us with \(\frac{2h}{7}\).
• Dividing by \(h\) simplifies to \(\frac{2}{7}\) after cancelling common factors.

Through simplification, we ensure clarity, paving the way to evaluate limits easily. With a clean expression, reaching the final answer is a more straightforward process, and we find the derivative of our function to be \(\frac{2}{7}\).