/*! 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 23 Use the formula $$f^{\prime}(x... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the formula $$f^{\prime}(x)=\lim _{z \rightarrow x} \frac{f(z)-f(x)}{z-x}$$ to find the derivative of the functions $$f(x)=\frac{1}{x+2}$$

Short Answer

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

Step by step solution

01

Set Up the Limit Definition for Derivative

We are given the function \( f(x) = \frac{1}{x+2} \). The derivative of the function \( f \) at a point \( x \) is given by the formula: \[ f^{\prime}(x)=\lim _{z \rightarrow x} \frac{f(z)-f(x)}{z-x} \] Substitute \( f(x) \) and \( f(z) \) into the formula: \[ f^{\prime}(x) = \lim_{z \to x} \frac{\frac{1}{z+2} - \frac{1}{x+2}}{z-x} \]
02

Simplify the Expression Inside the Limit

Find a common denominator for the terms inside the fraction: \[ \frac{1}{z+2} - \frac{1}{x+2} = \frac{x+2 - z-2}{(z+2)(x+2)} = \frac{x-z}{(z+2)(x+2)} \] Thus, the expression inside the limit becomes: \[ \frac{x-z}{(z+2)(x+2)(z-x)} \]
03

Simplify Using a Change of Variable

Rewrite \( x-z \) as \( -(z-x) \) to simplify the fraction: \[ \frac{x-z}{z-x} = -1 \] This simplifies the expression to: \[ \frac{-1}{(z+2)(x+2)} \]
04

Evaluate the Limit

Apply the limit as \( z \to x \): \[ \lim_{z \to x} \frac{-1}{(z+2)(x+2)} = \frac{-1}{(x+2)^2} \] Thus, the derivative of \( f(x) = \frac{1}{x+2} \) is \( f^{\prime}(x) = \frac{-1}{(x+2)^2} \).

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Key Concepts

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

Derivative Calculation
Calculating the derivative of a function involves finding the rate of change of the function's output with respect to its input. In simpler terms, it's like figuring out how much the function is bending or curving at any given point. For basic functions, you might already know rules for deriviatives. But for more complex functions, or when you want to understand the concept thoroughly, you use the limit definition of a derivative. Calculating derivatives can initially seem tricky, but once you understand the process, it becomes more intuitive. Always start by comprehensively understanding the function and apply the derivative rules accordingly. Calculating derivatives is a core skill in calculus, facilitating deeper understanding and more advanced analysis of functions. Breaking down the steps into manageable parts helps build confidence and improve problem-solving skills.
Limit Definition of Derivative
A key concept in calculus, the limit definition of a derivative provides the foundation for many differentiation techniques. The derivative of a function \( f(x) \) at a point \( x \) is elegantly expressed as: \[ f^{\prime}(x)=\lim _{z \rightarrow x} \frac{f(z)-f(x)}{z-x} \] This formula essentially measures the instantaneous rate of change of the function. It captures how the function behaves as you make very small changes in \( x \). In simpler terms, the derivative tells us the slope of the tangent line to the function at a given point. When applying the limit definition, begin by identifying your function \( f(x) \) and recognize that \( f(z) \) is the function at a nearby point \( z \). Substitute these into the definition, simplify the expression inside the limit, and then evaluate the limit as \( z \) approaches \( x \). This hands-on approach gives you detailed insights into how a function changes.
Rational Function Differentiation
When dealing with rational functions, which are ratios of polynomials, differentiation requires special attention. Rational functions, like our function \( f(x) = \frac{1}{x+2} \), involve more intricate operations. Differentiating such functions can sometimes be challenging due to their form. Using the limit definition in these scenarios often involves finding a common denominator and simplifying complex fractions.
  • Start by substituting the given function into the limit formula for the derivative.
  • Simplify the difference of functions to make the expression more manageable, typically by finding a common denominator.
  • Continue by simplifying the internal fraction, which might need factoring or rewriting parts for ease.
This guided process ensures precise calculation of the derivative and helps in breaking down potentially intimidating steps into understandable parts. For our example, the result is \( \frac{-1}{(x+2)^2} \), revealing the slope at any point \( x \) on the curve. This method hones your ability to handle more complex and varied calculus problems.

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Most popular questions from this chapter

Cardiac output In the late 1860 \(\mathrm{s}\) , Adolf Fick, a professor of physiology in the Faculty of Medicine in Wurzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 \(\mathrm{L} / \mathrm{min.}\) At rest it is likely to be a bit under 6 \(\mathrm{L} / \mathrm{min.}\) If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 \(\mathrm{L} / \mathrm{min} .\) $$\begin{array}{c}{\text { Your cardiac output can be calculated with the formula }} \\ {y=\frac{Q}{D}}\end{array}$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{ml} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{ml} / \mathrm{min}\) and \(D=97-56=41 \mathrm{ml} / \mathrm{L},\) $$y=\frac{233 \mathrm{ml} / \min }{41 \mathrm{ml} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min},$$ fairly close to the 6 \(\mathrm{L} / \mathrm{min}\) that most people have at basal (resting conditions. (Data courtesy of J. Kenneth Herd, M.D. Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

A sliding ladder A 13-ft ladder is leaning against a house when its base starts to slide away (see accompanying figure). By the time the base is 12 ft from the house, the base is moving at the rate of 5 \(\mathrm{ft} / \mathrm{sec}\). a. How fast is the top of the ladder sliding down the wall then? b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? c. At what rate is the angle \(\theta\) between the ladder and the ground changing then?

The radius of an inflating balloon A spherical balloon is inflated with helium at the rate of 100\(\pi \mathrm{ft}^{3} / \mathrm{min}\) . How fast is the balloon's radius increasing at the instant the radius is 5 \(\mathrm{ft} ?\) How fast is the surface area increasing?

The diameter of a sphere is measured as \(100 \pm 1 \mathrm{cm}\) and the volume is calculated from this measurement. Estimate the percentage error in the volume calculation.

Suppose that the functions \(f\) and \(g\) and their derivatives with respect to \(x\) have the following values at \(x=0\) and \(x=1 .\) $$\begin{array}{|c|c|c|c|}\hline x & {f(x)} & {g(x)} & {f^{\prime}(x)} & {g^{\prime}(x)} \\ \hline 0 & {1} & {1} & {5} & {1 / 3} \\ \hline 1 & {3} & {-4} & {-1 / 3} & {-8 / 3} \\ \hline\end{array}$$ Find the derivatives with respect to \(x\) of the following combinations at the given value of \(x .\) $$\begin{array}{ll}{\text { a. } 5 f(x)-g(x),} & {x=1 \quad \text { b. } f(x) g^{3}(x), \quad x=0} \\ {\text { c. } \frac{f(x)}{g(x)+1}, \quad x=1} & {\text { d. } f(g(x)), \quad x=0}\end{array}$$ $$\begin{array}{l}{\text { e. } g(f(x)), \quad x=0 \quad \text { f. }\left(x^{11}+f(x)\right)^{-2}, \quad x=1} \\ {\text { g. } f(x+g(x)), \quad x=0}\end{array}$$

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