/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Find the derivative by the limit... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 17

Find the derivative by the limit process. \(f(x)=\frac{1}{x-1}\)

Short Answer

Expert verified
With the limit process using the difference quotient, the derivative of the function \(f(x)=\frac{1}{x-1}\) is \(f'(x) = \frac{-1}{(x-1)^2}.\)

Step by step solution

01

Write down the original function

The original function is \(f(x)=\frac{1}{x-1}\).
02

Compute \[f(x+h)\]

To compute \[f(x+h)\], insert \(x+h\) in place of \(x\) in the original function: \[f(x+h)=\frac{1}{(x+h)-1} = \frac{1}{x+h-1}.\]
03

Compute the difference quotient

The difference quotient is the difference between \(f(x+h)\) and \(f(x)\) divided by \(h\). So it is \[\frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{x+h-1} - \frac{1}{x-1}}{h}.\]
04

Simplify the difference quotient

Here, the common denominator for the fractions is \[((x-1)(x+h-1))\]. Now we can rewrite the fraction as \[\frac{x-1-(x+h-1)}{h\cdot(x-1)(x+h-1)} = \frac{-h}{h\cdot(x-1)(x+h-1)} = \frac{-1}{(x-1)(x+h-1)}.\]
05

Take the limit as h approaches zero

This gives us the derivative: \[f'(x) = \lim_{h \rightarrow 0} \frac{-1}{(x-1)(x+h-1)} = \frac{-1}{(x-1)^2}.\] The limit process implies plugging \(h = 0\) into our difference quotient.

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Most popular questions from this chapter

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