/*! 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 43 1-44. Find the derivative of eac... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

1-44. Find the derivative of each function. $$ f(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} $$

Short Answer

Expert verified
Derivative: \( f'(x) = \frac{-4}{(e^x - e^{-x})^2} \)

Step by step solution

01

Identify the Rule to Use

Since the function is a rational function \( f(x) = \frac{e^x + e^{-x}}{e^x - e^{-x}} \), we need to use the Quotient Rule for differentiation. The Quotient Rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).
02

Differentiate the Numerator and Denominator

First, identify \( u(x) = e^x + e^{-x} \) and \( v(x) = e^x - e^{-x} \). Then find their derivatives: \[ u'(x) = \frac{d}{dx}(e^x + e^{-x}) = e^x - e^{-x} \]\[ v'(x) = \frac{d}{dx}(e^x - e^{-x}) = e^x + e^{-x} \]
03

Apply the Quotient Rule

Substitute \( u(x), u'(x), v(x), \) and \( v'(x) \) into the Quotient Rule formula:\[ f'(x) = \frac{(e^x - e^{-x})(e^x - e^{-x}) - (e^x + e^{-x})(e^x + e^{-x})}{(e^x - e^{-x})^2} \]
04

Simplify the Expression

Simplify the numerator:- Expand both terms: - \((e^x - e^{-x})(e^x - e^{-x}) = e^{2x} - 2 + e^{-2x}\) - \((e^x + e^{-x})(e^x + e^{-x}) = e^{2x} + 2 + e^{-2x}\)- The numerator becomes: \(e^{2x} - 2 + e^{-2x} - (e^{2x} + 2 + e^{-2x}) = -4\)
05

Simplify Further

Since the expression \( -4 \) is constant and not dependent on \( x \), simplify the derivative:The final derivative simplifies to:\[ f'(x) = \frac{-4}{(e^x - e^{-x})^2} \]
06

Conclusion

The derivative of the function \( f(x) = \frac{e^x + e^{-x}}{e^x - e^{-x}} \) is \( f'(x) = \frac{-4}{(e^x - e^{-x})^2} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Quotient Rule
In calculus, the Quotient Rule is a fundamental technique used for differentiating a specific type of functions known as rational functions, where one function is divided by another function. If you have a function \( f(x) = \frac{u(x)}{v(x)} \), the derivative \( f'(x) \) is given by:
  • Differentiate the numerator \( u(x) \) to get \( u'(x) \).
  • Differentiate the denominator \( v(x) \) to get \( v'(x) \).
  • Apply the formula: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]
This rule is especially useful when both the numerator and the denominator are non-constant functions, as it allows you to systematically find the derivative of the quotient. In our example, we used the Quotient Rule for the function \( f(x) = \frac{e^x + e^{-x}}{e^x - e^{-x}} \), which is a perfect case for this rule.
Exponential Functions
Exponential functions are functions of the form \( f(x) = a^{x} \) where \( a \) is a positive constant. In calculus, one of the most common exponential functions is the natural exponential function \( e^x \). The beauty of exponential functions like \( e^x \) is that their derivative is the same as the function itself: \( \frac{d}{dx}e^x = e^x \).
If you have expressions like \( e^{-x} \), you use the chain rule for differentiation, which involves multiplying by the derivative of the exponent: \( \frac{d}{dx}e^{-x} = -e^{-x} \).
These rules allow us to easily differentiate each part of the function when applying the Quotient Rule. For instance, in our exercise, we applied these rules to calculate the derivatives of the numerator and denominator.
Rational Functions
Rational functions are functions formed by dividing one polynomial by another. In our specific case, the rational function involves exponential terms rather than polynomial terms, which is not a typical scenario. However, for the purpose of differentiation using the Quotient Rule, they can be treated similarly since the technique applies to any functions as long as they are in a division form.
In such cases, identifying each part in the numerator and denominator correctly is crucial since each has unique derivatives. The function we worked with, \( f(x) = \frac{e^x + e^{-x}}{e^x - e^{-x}} \), is a great example of this.
After applying the Quotient Rule, we found the derivative to be a constant times a rational expression: \( f'(x) = \frac{-4}{(e^x - e^{-x})^2} \). Understanding how to manage the structure of rational functions, even with exponential components, is key to finding derivatives correctly.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that you have a bank account with interest compounded continuously, but you can't remember the continuously compounded interest rate. If at the end of the year you had \(10 \%\) more than you began with, was the continuously compounded rate more than \(10 \%\) or less than \(10 \% ?\)

BEHAVIORAL SCIENCE: Ebbinghaus Memory Model According to the Ebbinghaus model of memory, if one is shown a list of items, the percentage of items remembered \(t\) time units later is \(P(t)=(100-a) e^{-b t}+a\) where \(a\) and \(b\) are constants. For \(a=25\) and \(b=0.2\) this function becomes \(P(t)=75 e^{-0.2 t}+25 .\) Find the instantaneous rate of change of this percentage: a. at the beginning of the test \((t=0)\). b. after 3 time units.

BIOMEDICAL: DWI and Crash Risk The crash risk of an intoxicated driver relative to a similar driver with zero blood alcohol is $$ R(x)=\frac{195}{1+1650 e^{-35.5 x}} $$ where \(x\) is the blood alcohol level as a percent \((0.01 \leq x \leq 0.25) .\) For example, \(R(0.16)=29.4\) means that a driver with blood alcohol level \(0.16 \%\) is 29.4 times more likely to be involved in an accident than a similar driver who is not impaired. Find \(R^{\prime}(0.16)\) and interpret your answer. [Note: \(x \geq 0.08\) defines "driving while intoxicated."]

GENERAL: Population The United States population (in millions) is predicted to be \(P(t)=317 e^{0.01 t}\), where \(t\) is the number of years after \(2013 .\) Find the instantaneous rate of change of the population in the year \(2023 .\)

101-107. Choose the correct answer: $$ \frac{d}{d x} e^{x}=\quad \text { a. } x e^{x-1} \quad \text { b. } e^{x} \quad \text { c. } 0 $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.