Problem 55 - 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 \(\Delta x \rightarrow 0 .\) $$ y=\frac{x-1}{x+1} $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{2}{(x+1)^2}\)

Step 1: Find f(x+Δx)

First, we need to find what the function becomes when we replace \(x\) with \(x + \Delta x\) in \(y = \frac{x-1}{x+1}\). We have: \[ f(x + \Delta x) = \frac{(x + \Delta x) - 1}{(x + \Delta x) + 1} = \frac{x + \Delta x - 1}{x + \Delta x + 1} \]

Step 2: Set up \(\frac{\Delta y}{\Delta x}\)

Now use the expression for \(f(x + \Delta x)\) to find \(\frac{\Delta y}{\Delta x}\). Substitute \(f(x + \Delta x)\) and \(f(x)\) into the difference quotient:\[ \frac{\Delta y}{\Delta x} = \frac{\frac{x + \Delta x - 1}{x + \Delta x + 1} - \frac{x - 1}{x + 1}}{\Delta x} \]

Step 3: Simplify Fraction

To simplify, find a common denominator for the two fractions in the numerator. The common denominator is \[(x + \Delta x + 1)(x + 1)\].Calculate \(f(x + \Delta x) - f(x)\):\[ \frac{(x + \Delta x - 1)(x + 1) - (x - 1)(x + \Delta x + 1)}{(x + \Delta x + 1)(x + 1)} \]

Step 4: Simplify the Numerator

Expand and simplify the expression in the numerator:1. \((x + \Delta x - 1)(x + 1) = x^2 + x \Delta x - x + x + \Delta x - 1 = x^2 + x \Delta x + \Delta x - 1\)2. \((x - 1)(x + \Delta x + 1) = x^2 + x \Delta x + x - x - \Delta x - 1 = x^2 + x \Delta x - \Delta x - 1\)Subtract the second from the first expression:\[ x^2 + x \Delta x + \Delta x - 1 - (x^2 + x \Delta x - \Delta x - 1) \]Simplify to get:\[ 2\Delta x \]

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

Now substitute this back into \(\frac{\Delta y}{\Delta x}\):\[ \frac{\Delta y}{\Delta x} = \frac{2\Delta x}{\Delta x((x+\Delta x+1)(x+1))} \]Cancel \(\Delta x\) from numerator and denominator:\[ \frac{2}{(x + \Delta x + 1)(x + 1)} \]

Step 6: Take the Limit as \(\Delta x \rightarrow 0\)

Finally, find the derivative \(\frac{dy}{dx}\) by taking the limit of \(\frac{\Delta y}{\Delta x}\) as \(\Delta x\) approaches 0:\[ \frac{dy}{dx} = \lim_{\Delta x \to 0}\frac{2}{(x + \Delta x + 1)(x + 1)} \]This simplifies to:\[ \frac{dy}{dx} = \frac{2}{(x+1)^2} \]

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in calculus, specifically when discussing derivatives. At its core, the difference quotient is about measuring how a function changes as its input changes. Given a function \( f(x) \), the difference quotient is generally expressed as follows:\[ \frac{f(x+\Delta x)-f(x)}{\Delta x} \]This expression helps us understand the average rate of change of the function \( f \) over the interval from \( x \) to \( x + \Delta x \).

• The numerator \( f(x+\Delta x)-f(x) \) represents the change in the function's output.
• The denominator \( \Delta x \) represents the change in the input.

Think of the difference quotient as the slope of the secant line that passes through the points \((x, f(x))\) and \((x+\Delta x, f(x+\Delta x))\). In this context, calculating the difference quotient is the first step towards finding the derivative of the function, which captures the function's instantaneous rate of change. In the original exercise, we applied the difference quotient to the rational function \( y = \frac{x-1}{x+1} \) by substituting \( x + \Delta x \) into \( f \) and simplifying the resulting expression.

Limit Process for Derivatives

To transition from the average rate of change to the instantaneous rate of change, which is the derivative, we employ the limit process. The derivative of a function at a point is the limit of the difference quotient as \( \Delta x \) approaches 0. This is represented mathematically as:\[ \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} \]This limit converts the slope of the secant line into the slope of the tangent line, which accurately captures the function's behavior at an exact point.

• The key idea is that by making \( \Delta x \) infinitesimally small, the difference quotient approaches the derivative.
• In practice, you simplify the difference quotient as much as possible before applying the limit.

In our exercise, after simplifying the difference quotient for \( y = \frac{x-1}{x+1} \), taking the limit as \( \Delta x \rightarrow 0 \) directly gave us the derivative:\[ \frac{dy}{dx} = \frac{2}{(x+1)^2} \]This expression now describes the rate of change of \( y \) with respect to \( x \), considering the effects of the surrounding inputs.

Function Simplification Step-by-Step

Simplifying functions step-by-step is crucial when working with expressions involving difference quotients. Simplification makes it easier to see the underlying mechanics of complex expressions. For the given function \( y = \frac{x-1}{x+1} \), the simplification process involves several stages:

• Substitute \( x + \Delta x \) into the function: This step helps prepare the function for difference quotient application.
• Combine fractions: Calculate a common denominator to unite differing fractions. This reduces complexity and makes further operations easier.
• Simplify the numerator: Expand and combine like terms to reduce the expression to its simplest form. For example, in our exercise, this simplification resulted in the term \( 2\Delta x \).
• Eliminate \( \Delta x \): Cancel \( \Delta x \) from both the numerator and denominator to simplify before taking limits. In the denominator, any arithmetic errors are avoided by using correct algebraic manipulation.

This meticulous simplification ensures the remaining expression is manageable for taking limits, laying down a solid foundation for derivative calculation. By following these steps, complex rational functions can be differentiated with clarity and ease, avoiding pitfalls of algebraic errors or misinterpretation.