Problem 56 - Calculus Volume 1

Question

For the following exercises, 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 \( f'(x) = \frac{2}{7} \).

Step 1: Understand the Definition of a Derivative

The definition of the derivative of a function \( f(x) \) at a point \( x \) is given by the limit \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]. Our task is to apply this definition to find \( f'(x) \) for the function \( f(x) = \frac{2x}{7} + 1 \).

Step 2: Substitute into the Derivative Definition

Substitute \( f(x) = \frac{2x}{7} + 1 \) and \( f(x+h) = \frac{2(x+h)}{7} + 1 \) into the derivative formula. The subtraction \( f(x+h) - f(x) \) becomes: \[ \frac{2(x+h)}{7} + 1 - \left(\frac{2x}{7} + 1\right) \].

Step 3: Simplify the Expression

Simplify \( \frac{2(x+h)}{7} + 1 - \left(\frac{2x}{7} + 1\right) \). The 1s will cancel out, and the expression simplifies to: \[ \frac{2x + 2h}{7} - \frac{2x}{7} = \frac{2h}{7} \].

Step 4: Calculate the Limit

Now, calculate the limit: \( \lim_{h \to 0} \frac{\frac{2h}{7}}{h} \). This simplifies to \( \lim_{h \to 0} \frac{2}{7} \), which is just \( \frac{2}{7} \) since it does not depend on \( h \).

Step 5: State the Derivative

After evaluating the limit, we find that \( f'(x) = \frac{2}{7} \). This means that the derivative of the function \( f(x) = \frac{2x}{7} + 1 \) is a constant, \( \frac{2}{7} \).

Key Concepts

Definition of Derivative

The derivative of a function is a fundamental concept in calculus. It essentially tells us how a function changes as its input changes. To find the derivative of a function, we use the definition:

• The derivative of a function \( f(x) \) at a point \( x \) is expressed as: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] This formula is crucial for calculating the rate of change of the function at any given point.
• In our example, we apply this definition to the function \( f(x) = \frac{2x}{7} + 1 \).

Understanding this formula is key to mastering derivatives. It provides a systematic approach to finding the slope of the tangent line at any point on the function.

Limit Process

The limit process is an essential step in finding a derivative. It's the step where we see how the function behaves as its inputs change infinitesimally.

• For our function \( f(x) = \frac{2x}{7} + 1 \), we substitute the expressions \( f(x) \) and \( f(x+h) \) into the derivative formula.
• The subtraction \( f(x+h) - f(x) \) is simplified to \( \frac{2(x+h)}{7} + 1 - \left(\frac{2x}{7} + 1\right) \).

This results in the expression \( \frac{2h}{7} \) as the terms involving \( x \) and constant terms cancel out. The limit is then calculated as \( \lim_{h \to 0} \frac{\frac{2h}{7}}{h} \), which simplifies to \( \frac{2}{7} \). This shows how the difference in the function values becomes very small as \( h \) approaches zero.

Function Simplification

Function simplification is often necessary while working with derivatives to make calculations easier. This involves algebraically reducing complex expressions to simpler forms.

• In this exercise, after substituting the expressions into the derivative formula, simplifying \( \frac{2(x+h)}{7} + 1 - \left(\frac{2x}{7} + 1\right) \) results in \( \frac{2h}{7} \).
• This involves cancelling out the constant terms and terms involving \( x \).

Such simplifications not only make the expression easier to handle but also prepare it for applying the limit process. A clear and neat simplification ensures accuracy in calculating derivatives.

Constant Derivative

When calculating derivatives, sometimes you find a special case where the derivative is a constant value. This means that the rate of change of the function is uniform regardless of \( x \).

• For the function \( f(x) = \frac{2x}{7} + 1 \), after completing the limit process, we found that \( f'(x) \) is \( \frac{2}{7} \).
• This indicates a constant derivative, representing a straight-line slope for the linear function given.

A constant derivative tells us that the function's rate of change does not depend on \( x \), providing a straightforward understanding of linear functions in calculus. This makes constant derivatives a perfect example of uniform changes across an input range.