Problem 38 - College Algebra and Calculus: An Applied Approach

Question

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

Step-by-Step Solution

Answer

The derivative of the function \( g(s) = \frac{1}{s - 1} \) using the limit definition is \( g'(s) = -\frac{1}{(s-1)^2} \).

Step 1: Plug the function into the limit definition

Place the function \( g(s) = \frac{1}{s-1} \) into the limit definition which gives the following: \( g'(s) = \lim_{h \to 0} \frac{ \frac{1}{s+h-1} - \frac{1}{s-1}}{h} \).

Step 2: Simplify the fraction in the numerator

It's necessary to simplify the term in the numerator of the fraction which can be done using the formula for the difference of fractions: \( \frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd} \). Apply this formula and re-write the term in the numerator: \( \frac{ (s-1) - (s+h-1)}{(s+h-1)(s-1)} \). After simplification, the result will be \( -\frac{h}{(s+h-1)(s-1)} \).

Step 3: Simplify the entire expression

The derivative will be: \( g'(s) = \lim_{h \to 0} \frac{-\frac{h}{(s+h-1)(s-1)}}{h} \). Cancelling \( h \) on the numerator and the denominator leaves: \( g'(s) = -\lim_{h \to 0} \frac{1}{(s+h-1)(s-1)} \).

Step 4: Evaluate the limit

Applying the limit gives the derivative: \( g'(s) = -\frac{1}{(s-1)^2} \). This is the derivative of the function \( g(s) = \frac{1}{s - 1} \).