Problem 56 - Calculus Early Transcendentals: Pearson New International Edition

Question

First find and simplify $$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ Then find \(d y / d x\) by taking the limit of your answer as \(\Delta x \rightarrow 0 .\) $$ y=\frac{x^{2}-1}{x} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{2}{x} \)

Step 1: Identify the function

The function given is \( y = \frac{x^2 - 1}{x} \). This is a rational function.

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

The function after replacing \( x \) with \( x + \Delta x \) is \( f(x + \Delta x) = \frac{(x + \Delta x)^2 - 1}{x + \Delta x} \).

Step 3: Expand and simplify \( f(x + \Delta x) \)

Expand \((x + \Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2 \). So, \( f(x + \Delta x) = \frac{x^2 + 2x\Delta x + (\Delta x)^2 - 1}{x + \Delta x} \).

Step 4: Calculate \( \Delta y \)

\( \Delta y = f(x + \Delta x) - f(x) = \frac{x^2 + 2x\Delta x + (\Delta x)^2 - 1}{x + \Delta x} - \frac{x^2 - 1}{x} \).

Step 5: Simplify \( \frac{\Delta y}{\Delta x} \)

Find a common denominator, which is \( x(x + \Delta x) \), and simplify: \[ \frac{\Delta y}{\Delta x} = \frac{\left( x^2 + 2x\Delta x + (\Delta x)^2 - 1 \right) x - \left( x^2 - 1 \right)(x + \Delta x)}{\Delta x \cdot x(x+\Delta x)} \]. Expand, simplify the numerator, and factor out \( \Delta x \).

Step 6: Simplify further

The simplified form for the numerator becomes \( 2x\Delta x + (\Delta x)^2 \), so the expression is \( \frac{\Delta y}{\Delta x} = \frac{2x + \Delta x}{x(x + \Delta x)} \) after cancelling out \( \Delta x \).

Step 7: Take the Limit as \( \Delta x \rightarrow 0 \)

Taking the limit as \( \Delta x \rightarrow 0 \), the term \( \Delta x \) in the numerator goes to 0, leading to: \[ \lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} = \frac{2x}{x^2} \], which simplifies to \( \frac{2}{x} \).

Step 8: Final Result: Determine \( \frac{dy}{dx} \)

The derivative of the function \( y \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{2}{x} \).

Key Concepts

Derivative of a Rational Function

Understanding how to find the derivative of a rational function is essential in calculus. A rational function, like the one given in our exercise, is a fraction comprised of two polynomials. Our function is \( y = \frac{x^2 - 1}{x} \). To determine its derivative, we need to use the definition of the derivative, which involves calculating \( \frac{\Delta y}{\Delta x} \). This means we find the difference in the function values and divide it by the change in \( x \).
To start, our task is to compute \( f(x + \Delta x) \). We substitute \( x + \Delta x \) into the original function, transforming it into \( f(x + \Delta x) = \frac{(x + \Delta x)^2 - 1}{x + \Delta x} \). We then expand and simplify the expression to facilitate further steps. Understanding this process forms the foundation of calculating derivatives.

Limit Process in Calculus

In calculus, the limit process is a fundamental tool to transition from average rates of change to instantaneous rates of change, otherwise known as derivatives. To compute the derivative from \( \frac{\Delta y}{\Delta x} \), we need to approach this expression as \( \Delta x \) goes to 0.
After simplifying, our expression becomes \( \frac{2x + \Delta x}{x(x + \Delta x)} \). We identify that it still includes \( \Delta x \) in the numerator. To find the derivative, we take the limit as \( \Delta x \rightarrow 0 \), allowing us to handle the expression as if \( \Delta x \) is almost negligible. This results in \( \lim_{\Delta x \rightarrow 0} \frac{2x + \Delta x}{x(x + \Delta x)} = \frac{2x}{x^2} \), which simplifies further to \( \frac{2}{x} \).
By comprehending this limit process, students can grasp how derivatives represent the "speed" at which function values change, providing insights into the function's behavior at an exact point.

Simplification of Algebraic Expressions

Simplification is a critical strategy when dealing with complex algebraic expressions, especially in calculus. Simplifying terms in functions allows for clearer insights and more manageable calculations. Throughout this exercise, we encountered the necessity to expand polynomial expressions, which is crucial in obtaining a simplified derivative.
Initially, \( (x + \Delta x)^2 \) expanded to \( x^2 + 2x\Delta x + (\Delta x)^2 \). Instead of dealing with cumbersome expressions, we applied algebraic techniques to streamline our work.
Most prominent was the step \( \frac{2x\Delta x + (\Delta x)^2}{\Delta x \cdot x(x+\Delta x)} \). Here, factoring out \( \Delta x \) from the numerator substantially simplified the expression. This allowed cancellation of \( \Delta x \) from both numerator and denominator, leading to \( \frac{2x + \Delta x}{x(x + \Delta x)} \).
Recognizing opportunities for simplification enables cleaner calculations and reduces chances for error, making tasks much more approachable for students tackling calculus problems.