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Chapter 2: Problem 38

Use the limit definition to find the derivative of the function. $$ g(s)=\frac{1}{s-1} $$

Short Answer

Expert verified
The derivative of the function \( g(s)=\frac{1}{s-1} \) is \( - \frac{1}{{(s-1)^2}} \).

Step by step solution

01

Write the Function in the Difference Quotient

Substitute \( g(s+\Delta s)=\frac{1}{{s+\Delta s-1}} \) and \( g(s)=\frac{1}{s-1} \) into the limit definition \(\frac{g(s+\Delta s)-g(s)}{\Delta s}\), which results in \( \frac{\frac{1}{{s+\Delta s-1}}-\frac{1}{s-1}}{\Delta s} \).
02

Simplify the Difference Quotient

Combine the fractions on the numerator. This results in \( \frac{{s-1-(s+\Delta s-1)}}{(s+\Delta s-1)(s-1)\Delta s} = \frac{-\Delta s}{{(s+\Delta s-1)(s-1)\Delta s}} \). Then, cancelling out \(\Delta s\), we have \( - \frac{1}{{(s+\Delta s-1)(s-1)}} \).
03

Taking the Limit

To find the derivative, take the limit of the difference quotient as \(\Delta s\) approaches 0: \( lim_{\Delta s \to 0} - \frac{1}{{(s+\Delta s-1)(s-1)}} = - \frac{1}{{(s-1)^2}} \). This is the derivative of the function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Definition
The limit definition of a derivative is a fundamental concept in calculus. It provides a precise way to calculate the derivative of a function at any point. It's essentially about finding the instantaneous rate of change of the function, or the slope of the tangent line at a given point.
This is expressed mathematically as:
  • The derivative of a function \( f(x) \) at a point \( x \) is given by the limit:
  • \( \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} \)
For the function \( g(s) = \frac{1}{s-1} \), we use this definition to find its derivative. The process begins by substituting \( g(s+\Delta s) \) and \( g(s) \) into the difference quotient, which leads us to the next critical step.
Difference Quotient
The difference quotient is an essential component of the limit definition. It represents the average rate of change of the function over the interval \( \Delta s \). It's calculated as follows:
  • \( \frac{g(s+\Delta s) - g(s)}{\Delta s} \)
In our exercise, this becomes:
  • \( \frac{\frac{1}{s+\Delta s-1} - \frac{1}{s-1}}{\Delta s} \)
This step involves the challenge of simplifying complex fractions. Here, we need to find a common denominator for the fractions in the numerator. Upon simplification, we find that:
  • \( - \frac{1}{(s+\Delta s-1)(s-1)} \)
Simplifying the difference quotient prepares us to take the limit.
Derivative Calculation
Calculating the derivative involves taking the limit of the simplified difference quotient as \( \Delta s \) approaches zero. This step reveals the instantaneous rate of change of the function, giving us the derivative.
  • \( \lim_{\Delta s \to 0} - \frac{1}{(s+\Delta s-1)(s-1)} \)
In this specific exercise, as \( \Delta s \) becomes exceedingly small, the expression simplifies further. The term \( s + \Delta s - 1 \) becomes \( s - 1 \), leading to:
  • \( - \frac{1}{(s-1)^2} \)
Thus, the derivative \( g'(s) \) of the function \( g(s) = \frac{1}{s-1} \) is \( - \frac{1}{(s-1)^2} \), capturing the function's instantaneous rate of change at any point \( s \) not equal to 1. This is why reliably working through the limit and difference quotient is so critical in calculus.

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