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Chapter 7: Problem 37

Use the limit definition to find the derivative of the function. $$ f(x)=\frac{1}{x+2} $$

Short Answer

Expert verified
The derivative of the function \(f(x)=\frac{1}{x+2}\) is \(f'(x)=\frac{-1}{(x+2)^2}\).

Step by step solution

01

Set Up Limit Definition

Using the limit definition of a derivative, start by writing the setup equation: \(\lim_{h\to0} \frac{f(x+h)-f(x)}{h}\). This will be used to find the derivative for the function \(f(x)=\frac{1}{x+2}\).
02

Substitute Function

Substitute the function \(f(x)=\frac{1}{x+2}\) into the setup equation from Step 1. This results in: \(\lim_{h\to0} \frac{\frac{1}{x+h+2}-\frac{1}{x+2}}{h}\).
03

Simplify Difference Quotient

Simplify the difference quotient before taking the limit as h approaches 0. Multiply the numerator and the denominator by \((x+h+2)(x+2)\) to eliminate the complex fractions. This simplifies the equation to: \(\lim_{h\to0} \frac{(x+2)-(x+h+2)}{h(x+h+2)(x+2)}\). Next, simplify the numerator and the equation simplifies to: \(\lim_{h\to0} \frac{-h}{h(x+h+2)(x+2)}\). The h in the numerator and denominator cancel out, resulting in: \(\lim_{h\to0} \frac{-1}{(x+h+2)(x+2)}\).
04

Compute Limit

Substitute \(h=0\) into the equation now that the difference quotient is simplified. This will let us determine the actual limit: \( \lim_{h\to0} \frac{-1}{(x+2+0)(x+2)} = \frac{-1}{(x+2)^2}\).

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