Chapter 2: Problem 55
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 \(\Delta x \rightarrow 0 .\) $$ y=\frac{x-1}{x+1} $$
Short Answer
Expert verified
The derivative \( \frac{dy}{dx} \) is \( \frac{-2}{(x+1)^2} \).
Step by step solution
01
Define Function Components
Given the function \( y = \frac{x-1}{x+1} \), we need to evaluate the change in \( y \), represented by \( f(x+\Delta x) \) and \( f(x) \). Here, define \( f(x) = \frac{x-1}{x+1} \).
02
Substitute in Function
Let's compute \( f(x+\Delta x) \). Substitute \( x + \Delta x \) into the function: \[ f(x + \Delta x) = \frac{(x + \Delta x) - 1}{(x + \Delta x) + 1} = \frac{x + \Delta x - 1}{x + \Delta x + 1} \].
03
Compute \( \Delta y/\Delta x \)
Now, apply the expression for \( \Delta y/\Delta x \): \[ \frac{\Delta y}{\Delta x} = \frac{f(x+\Delta x) - f(x)}{\Delta x} = \frac{\frac{x + \Delta x - 1}{x + \Delta x + 1} - \frac{x - 1}{x + 1}}{\Delta x} \].
04
Simplify the Expression
Simplify by finding a common denominator for both fractions: \[ \frac{(x + \Delta x - 1)(x + 1) - (x - 1)(x + \Delta x + 1)}{(x+\Delta x + 1)(x + 1)\Delta x} \].
05
Expand and Combine Terms
Expand and simplify the numerator: \[ \frac{((x + \Delta x - 1)(x + 1)) - ((x-1)(x + \Delta x + 1))}{(x+\Delta x + 1)(x + 1)\Delta x} = \frac{(x^2 + x\Delta x + x - 1\Delta x - 1) - (x^2 + x\Delta x + 1 - x\Delta x - \Delta x - 1)}{(x+\Delta x + 1)(x + 1)\Delta x} \].
06
Cancel and Simplify Terms
Simplify the expression: \[ \frac{(x^2 + x\Delta x + x - \Delta x - 1) - (x^2 - 1(1))}{(x+\Delta x + 1)(x + 1)\Delta x} = \frac{-2\Delta x}{(x+\Delta x + 1)(x + 1)\Delta x} \]. Now cancel \(\Delta x\) in the numerator and denominator: \[ \frac{-2}{(x + \Delta x + 1)(x + 1)} \].
07
Compute the Limit as \( \Delta x \rightarrow 0 \)
To find the derivative \( \frac{dy}{dx} \), take the limit as \( \Delta x \to 0 \): \[ \lim_{\Delta x \to 0} \frac{-2}{(x + \Delta x + 1)(x + 1)} = \frac{-2}{(x+1)^2} \].
08
Conclusion
Thus, the derivative \( \frac{dy}{dx} \) is \( \frac{-2}{(x+1)^2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference Quotient
The difference quotient is a crucial concept in calculus that provides a way to approximate the slope of the tangent line to a function at a point. It helps in understanding how a function behaves over small intervals. The difference quotient formula is \( \frac{f(x + \Delta x) - f(x)}{\Delta x} \), where \( \Delta x \) represents a small change in \( x \).
The goal is to prepare for taking the limit by having a manageable algebraic expression.
- The numerator \( f(x + \Delta x) - f(x) \) provides the change in the function's output, \( y \), over the interval \( \Delta x \).
- The denominator \( \Delta x \) is the change in the input, \( x \).
The goal is to prepare for taking the limit by having a manageable algebraic expression.
Limit Process
The limit process is essential in transforming the difference quotient into the derivative, which gives the exact rate of change at a point. To find \( \frac{dy}{dx} \), you take the limit of the difference quotient as \( \Delta x \to 0 \). This mathematical technique refines our understanding of behavior at infinitesimally small scales.
- In the given exercise, after simplifying the difference quotient, it was necessary to substitute \( \Delta x = 0 \) into the expression \( \frac{-2}{(x + \Delta x + 1)(x + 1)} \).
- As \( \Delta x \to 0 \), this simplification results in \( \frac{-2}{(x+1)^2} \).
Rational Function Differentiation
Rational functions, like \( y = \frac{x-1}{x+1} \), are functions that can be expressed as the ratio of two polynomials. Differentiating such functions involves the quotient rule or simplifying into a form amenable to the difference quotient and limit process.
The key steps in dealing with rational functions include:
The key steps in dealing with rational functions include:
- Finding common denominators, especially when subtracting fractions, which is often necessary in forming the difference quotient.
- Simplifying the resulting expression so that terms cancel, leaving a clear path to taking the limit.