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Chapter 3: Problem 55

Let \(f(x)=x /\left(1+x^{2}\right) .\) Use the identity $$ \frac{x /\left(1+x^{2}\right)-c /\left(1+c^{2}\right)}{x-c}=\frac{1-x c}{\left(1+x^{2}\right)\left(1+c^{2}\right)} $$ to compute \(f^{\prime}(c)\).

Short Answer

Expert verified
The derivative \(f'(c) = \frac{1 - c^2}{(1 + c^2)^2}\).

Step by step solution

01

Understand the Given Identity

We are given an identity: \[\frac{x /(1+x^{2})-c /(1+c^{2})}{x-c}=\frac{1-x c}{(1+x^{2})(1+c^{2})}\] This is a specific setup to find the derivative of the function \(f(x)\) at \(x=c\). The identity simplifies the expression for the difference quotient, which is useful for evaluating the derivative.
02

Recall the Definition of the Derivative

The derivative of \(f(x)\) at a point \(c\) is defined as: \[f'(c) = \lim_{x \to c}\frac{f(x) - f(c)}{x - c}\] In this context, the equation from the problem provides the simplified form of the difference quotient for \(f(x) = x/(1 + x^2)\). Our job is to find \(\lim_{x \to c}\) of this expression.
03

Substitute and Simplify

Plug the given identity into the derivative's limit formula:\[f'(c) = \lim_{x \to c}\frac{x/(1+x^2)-c/(1+c^2)}{x-c}\]From the problem, we know:\[\frac{x/(1+x^2)-c/(1+c^2)}{x-c} = \frac{1-xc}{(1+x^2)(1+c^2)}\] Thus:\[f'(c) = \lim_{x \to c}\frac{1-xc}{(1+x^2)(1+c^2)}\]
04

Evaluate the Limit

Now evaluate the limit:\[\lim_{x \to c}\frac{1-xc}{(1+x^2)(1+c^2)}\]Substitute \(x = c\) directly since the expression is no longer in an indeterminate form:\[= \frac{1 - c^2}{(1 + c^2)(1 + c^2)} = \frac{1 - c^2}{(1 + c^2)^2}\] Thus, the derivative of the function \(f(x)\) at \(x = c\) is \(f'(c) = \frac{1 - c^2}{(1 + c^2)^2}\).

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Key Concepts

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

Difference Quotient
The concept of the difference quotient is central to understanding derivatives. It allows us to compute the derivative by approximating the slope of the tangent line to a curve at a point. The difference quotient formula is given by: \[\frac{f(x) - f(c)}{x - c}\] This represents the average rate of change of the function \(f(x)\) between \(x\) and \(c\).
  • In the context of the problem, the difference quotient we use is specific to the function \(f(x) = \frac{x}{1+x^2}\).
  • This quotient becomes simpler with provided identities, making the derivative calculation more straightforward.
Understanding this concept is key as it serves as a building block for finding the limit to define the derivative.
Definition of the Derivative
The derivative of a function at a point measures how the function changes as its input changes. For a function \(f(x)\), the derivative at a point \(c\) is expressed as:\[f'(c) = \lim_{x \to c}\frac{f(x) - f(c)}{x - c}\]
  • This expression means that the derivative is the limit of the difference quotient as \(x\) approaches \(c\).
  • It is about finding the instantaneous rate of change or the slope of the tangent at that point.
In our problem, we substitute the identity we have for the difference quotient, focusing on simplifying and evaluating the limit to find the derivative.
Limit
Limits are fundamental in calculus. They help us understand the behavior of functions as they approach a particular point. The limit process is involved in defining derivatives and handling indeterminate forms. For a function, the limit as \(x\) approaches a specific value gives insight into the function's behavior around that point.
  • In our problem, the limit \(\lim_{x \to c}\frac{1-xc}{(1+x^2)(1+c^2)}\) is evaluated when \(x\) approaches \(c\).
  • Once eliminated from indeterminacy, this allows us to plug in \(x = c\) directly to find the derivative.
Understanding limits is essential in calculus as they outline the foundation for derivative calculations.
Function
Functions describe relationships where each input is associated with exactly one output. In calculus, functions allow us to model a wide range of phenomena. The function \(f(x) = \frac{x}{1+x^2}\) captures a specific mathematical relationship.
  • This mathematical expression forms the foundation for building the derivative and limits that we examine in the problem.
  • Understanding the structure of the function allows for substitutions and comparisons needed for derivative evaluation.
Recognizing how functions behave helps us in manipulating and applying rules like the difference quotient to find solutions effectively.

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Most popular questions from this chapter

A demand curve is given. Use the method of implicit differentiation to find \(d q / d p .\) For the given price \(p_{0}\), solve the demand equation to find the corresponding demand \(q_{0}\). Then use the differential approximation with base point \(p_{0}\) to estimate the demand at price \(p_{1} .\) Find the exact demand at price \(p_{1} .\) What is the relative error that results from the differential approximation? $$ p^{2} q / 10+5 p q^{1 / 5}=1280200, p_{0}=1.80, p_{1}=2 $$

An initial value problem is given, along with its exact solution. (Read the instructions for Exercises \(47-50\) for terminology.) Verify that the given solution is correct by substituting it into the given differential equation and the initial value condition. Calculate the Euler's Method approximation \(y_{1}=y_{0}+F\left(x_{0}, y_{0}\right) \Delta x\) of \(y\left(x_{1}\right)\) where \(\Delta x=x_{1}-x_{0} .\) Let \(m_{1}=\left(F\left(x_{0}, y_{0}\right)+F\left(x_{1}, y_{1}\right)\right) / 2\) and \(z_{1}=y_{0}+\) \(m_{1} \Delta x .\) This is the Improved Euler Method approximation of \(y\left(x_{1}\right) .\) Calculate \(z_{1} .\) By evaluating \(y\left(x_{1}\right),\) determine which of the two approximations, \(y_{1}\) or \(z_{1},\) is more accurate. $$ \begin{array}{l} d y / d x=x^{2}-2 y, y(0)=3, x_{1}=1 / 4 ; \text { Exact solution: } y(x)= \\ x^{2} / 2-x / 2+1 / 4+11 / 4 \exp (-2 x) \end{array} $$

Calculate the linearization \(L(x)=f(c)+\) \(f^{\prime}(c), \cdot(x-c)\) for the given function \(f\) at the given value \(c\) $$ f(x)=\cos ^{2}(x), c=\pi / 4 $$

A function \(f,\) a point \(c,\) an increment \(\Delta x,\) and a positive integer \(n\) are given. Use the method of increments to estimate \(f(c+\Delta x)\). Then let \(h=\Delta x\) / \(N\). Use the method of increments to obtain an estimate \(y_{1}\) of \(f(c+h) .\) Now, with \(c+h\) as the base point and \(y_{1}\) as the value of \(f(c+h),\) use the method of increments to obtain an estimate \(y_{2}\) of \(f(c+2 h)\). Continue this process until you obtain an estimate \(y_{N}\) of \(f(c+N \cdot h)=f(c+\Delta x) .\) We say that we have taken \(N\) steps to obtain the approximation. The number \(h\) is said to be the step size. Use a calculator or computer to evaluate \(f(c+\Delta x)\) directly. Compare the accuracy of the one step and \(N\) -step approximations. $$ f(x)=\sqrt{x}, c=4, \Delta x=0.5, N=5 $$

We call this Euler's Method of approximating the unknown function \(y .\) In each of Exercises \(47-50,\) an initial value problem is given. Calculate the Euler's Method approximation \(y_{1}\) of \(y\left(x_{1}\right)\) $$ d y / d x=1+y / x, y(2)=1 / 2, x_{1}=3 / 2 $$
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