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Chapter 1: Problem 83

Use a derivative routine to obtain the value of the derivative. Give the value to 5 decimal places. $$f^{\prime}(2), \text { where } f(x)=\frac{x}{1+x}$$

Short Answer

Expert verified
The derivative \(f'(2)\) is approximately 0.11111.

Step by step solution

01

Identify the Function

The given function is written as \( f(x) = \frac{x}{1+x} \). To find the derivative of this function at a specific point, we first need to differentiate the function.
02

Apply the Quotient Rule

The quotient rule for differentiation states that if \( f(x) = \frac{u(x)}{v(x)} \), then \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \]. Here, \( u(x) = x \) and \( v(x) = 1 + x \).
03

Differentiate the Numerator and the Denominator

The derivative of the numerator \( u(x) = x \) is \( u'(x) = 1 \). The derivative of the denominator \( v(x) = 1 + x \) is \( v'(x) = 1 \).
04

Putting It All Together

Substitute the derivatives and the original functions into the quotient rule formula: \[ f'(x) = \frac{1(1+x) - x(1)}{(1+x)^2} = \frac{1 + x - x}{(1 + x)^2} = \frac{1}{(1 + x)^2} \].
05

Evaluate at x = 2

Substitute \( x = 2 \) into the derivative function: \[ f'(2) = \frac{1}{(1 + 2)^2} = \frac{1}{3^2} = \frac{1}{9} \].
06

Final Value to Five Decimal Places

Calculate \( \frac{1}{9} \) to five decimal places: \( f'(2) \approx 0.11111 \).

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Key Concepts

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

quotient rule
The quotient rule is essential when differentiating a function that is a ratio of two smaller functions. It's like having a tool to deal with fractions but in calculus.
The rule states:
If you have a function defined as \( f(x) = \frac{u(x)}{v(x)} \), you can find its derivative using:
\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \].
  • First, identify the functions \( u(x) \) and \( v(x) \).
  • Second, find their respective derivatives, \( u'(x) \) and \( v'(x) \).
  • Third, plug these values into the quotient rule formula.
In our given problem, we applied the quotient rule to the function \( f(x) = \frac{x}{1+x} \). We identified \( u(x) = x \) and \( v(x) = 1 + x \). Their derivatives are \( u'(x) = 1 \) and \( v'(x) = 1 \).
This simplifies into: \[ f'(x) = \frac{1(1+x) - x(1)}{(1+x)^2} = \frac{1 + x - x}{(1 + x)^2} = \frac{1}{(1 + x)^2} \].
differentiation
Differentiation is the process of finding the derivative of a function. It tells you how the function's value changes as its input changes, which is crucial in understanding the behavior of functional relationships.
To differentiate a function means to find its rate of change. The quotient rule is one of several techniques used in differentiation.
Let's break down some key points:
  • When you differentiate a constant, you get zero.
  • Differentiating \( x^n \) gives \( n x^{n-1} \).
  • For our problem, the initial function was \( f(x) = \frac{x}{1+x} \), which required the quotient rule.
By following a systematic differentiation strategy, you can simplify complex functions into more workable forms. In our case, after applying differentiation to \( f(x) \), we found:
\[ f'(x) = \frac{1}{(1 + x)^2} \].
numerical approximation
Numerical approximation involves getting a value close to an exact number. It鈥檚 useful when an exact value is complex to compute.
In our problem, once we found the derivative function \( f'(x) = \frac{1}{(1 + x)^2} \), we needed to evaluate it at \( x = 2 \).
Substituting \( x = 2 \), we obtained:
\[ f'(2) = \frac{1}{(3)^2} = \frac{1}{9} \].
To provide a more useful answer, approximate \( \frac{1}{9} \) to five decimal places:
\[ f'(2) \approx 0.11111 \].
This gives a more practical insight into how the function changes at that specific point. Numerical approximations are heavily used in calculus when exact forms are either unnecessary or hard to express accurately.

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Most popular questions from this chapter

In a psychology experiment, people improved their ability to recognize common verbal and semantic information with practice. Their judgment time after \(t\) days of practice was \(f(t)=.36+.77(t-.5)^{-0.36}\) seconds. (Source: American Journal of Psychology.) (a) Display the graphs of \(f(t)\) and \(f^{\prime}(t)\) in the window \([.5,6]\) by \([-3,3]\). Use these graphs to answer the following questions. (b) What was the judgment time after 4 days of practice? (c) After how many days of practice was the judgment time about .8 second? (d) After 4 days of practice, at what rate was judgment time changing with respect to days of practice? (e) After how many days was judgment time changing at the rate of \(-.08\) second per day of practice?

Consider the cost function \(C(x)=6 x^{2}+14 x+18\) (thousand dollars). (a) What is the marginal cost at production level \(x=5 ?\) (b) Estimate the cost of raising the production level from \(x=5\) to \(x=5.25.\) (c) Let \(R(x)=-x^{2}+37 x+38\) denote the revenue in thousands of dollars generated from the production of \(x\) units. What is the breakeven point? (Recall that the breakeven point is when revenue is equal to cost.) (d) Compute and compare the marginal revenue and marginal cost at the breakeven point. Should the company increase production beyond the breakeven point? Justify your answer using marginals.

Use limits to compute \(f^{\prime}(x) .\) $$f(x)=\frac{1}{x^{2}+1}$$

Determine which of the following limits exist. Compute the limits that exist. $$\lim _{x \rightarrow 8} \frac{x^{2}+64}{x-8}$$

Use limits to compute the following derivatives. $$f^{\prime}(0), \text { where } f(x)=x^{3}+3 x+1$$
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