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Chapter 6: Problem 50

Find the derivatives of the following functions. $$f(x)=\operatorname{csch}^{-1}(2 / x)$$

Short Answer

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Question: Find the derivative of the function $$f(x)=\operatorname{csch}^{-1}\left(\frac{2}{x}\right)$$. Answer: The derivative of the function $$f(x)=\operatorname{csch}^{-1}\left(\frac{2}{x}\right)$$ is $$\frac{df}{dx} = \frac{x}{\sqrt{x^2+4}}$$.

Step by step solution

01

Find the derivative of the argument with respect to $$x$$

We have $$u = \frac{2}{x}$$. Let's find $$\frac{du}{dx}$$. Using the quotient rule, we get: $$\frac{du}{dx} = \frac{-2}{x^2}$$
02

Find the derivative of $$\operatorname{csch}^{-1}$$ with respect to $$u$$

The derivative of the inverse hyperbolic cosecant function is given by: $$\frac{d}{du}\operatorname{csch}^{-1}(u) = -\frac{1}{u\sqrt{1+u^2}}$$
03

Apply the chain rule

Now we apply the chain rule, multiplying the results of Step 1 and Step 2: $$\frac{df}{dx} = \frac{d}{dx}\operatorname{csch}^{-1}\left(\frac{2}{x}\right) = -\frac{1}{u\sqrt{1+u^2}} \cdot \frac{-2}{x^2}$$ Substitute $$u = \frac{2}{x}$$ back into the expression: $$\frac{df}{dx} = -\frac{1}{(\frac{2}{x})\sqrt{1+(\frac{2}{x})^2}} \cdot \frac{-2}{x^2}$$
04

Simplify the expression

Simplify the expression to get the final answer: $$\frac{df}{dx} = \frac{1 \cdot 2}{(\frac{2}{x})\sqrt{1+(\frac{2}{x})^2} \cdot x^2}$$ $$\frac{df}{dx} = \frac{2x}{2\sqrt{x^2+4}}$$ $$\boxed{\frac{df}{dx} = \frac{x}{\sqrt{x^2+4}}}$$ The derivative of the function $$f(x)=\operatorname{csch}^{-1}\left(\frac{2}{x}\right)$$ is $$\frac{df}{dx} = \frac{x}{\sqrt{x^2+4}}$$.

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Key Concepts

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

Derivative
The concept of a derivative involves finding the rate at which a function changes at any given point. It is a fundamental idea in calculus, and it helps us understand how functions behave.
For any function \( f(x) \), the derivative \( \frac{df}{dx} \) measures how \( f \) changes as \( x \) changes.
  • The process is often visualized as finding the slope of the tangent line to the curve at a particular point.
  • Derivatives can be used to find the maximum or minimum values of functions, solve real-world problems, and model physical systems.
In the given exercise, we applied differentiation to \( f(x)=\operatorname{csch}^{-1}(2 / x) \) using standard differentiation rules.
Inverse Hyperbolic Functions
Inverse hyperbolic functions are the inverses of the hyperbolic functions like \( \operatorname{csch}(x) \), \( \operatorname{sinh}(x) \), etc.
These functions are useful in various fields, including engineering and physics, as they help model some natural phenomena.
  • The function \( \operatorname{csch}^{-1}(x) \) is the inverse hyperbolic cosecant.
  • They have specific properties and formulas for differentiation, which are crucial in calculus.
In the exercise, we used the formula for the derivative of \( \operatorname{csch}^{-1}(u) \) which is \( -\frac{1}{u\sqrt{1+u^2}} \). This formula helps in finding how inverse hyperbolic functions change.
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions.
This rule allows us to differentiate functions that are made up of other functions.
  • The general form of the chain rule is \( \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \).
  • It's particularly useful when combining different types of functions, such as trigonometric, exponential, or logarithmic functions.
In this exercise, we applied the chain rule to the function \( \operatorname{csch}^{-1}(2 / x) \).
First, we found the derivative of \( u = \frac{2}{x} \), and then used the derivative of \( \operatorname{csch}^{-1}(u) \) to calculate the overall derivative.

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Most popular questions from this chapter

The burning of fossil fuels releases greenhouse gases (roughly \(60 \% \text { carbon dioxide })\) into the atmosphere. In 2010 , the United States released approximately 5.8 billion metric tons of carbon dioxide (Environmental Protection Agency estimate), while China released approximately 8.2 billion metric tons (U.S. Department of Energy estimate). Reasonable estimates of the growth rate in carbon dioxide emissions are \(4 \%\) per year for the United States and \(9 \%\) per year for China. In 2010 , the U.S. population was 309 million, growing at a rate of \(0.7 \%\) per year, and the population of China was 1.3 billion, growing at a rate of \(0.5 \%\) per year. a. Find exponential growth functions for the amount of carbon dioxide released by the United States and China. Let \(t=0\) correspond to 2010 . b. According to the models in part (a), when will Chinese emissions double those of the United States? c. What was the amount of carbon dioxide released by the United States and China per capita in \(2010 ?\) d. Find exponential growth functions for the per capita amount of carbon dioxide released by the United States and China. Let \(t=0\) correspond to 2010. e. Use the models of part (d) to determine the year in which per capita emissions in the two countries are equal.

Show that \(\cosh ^{-1}(\cosh x)=|x|\) by using the formula \(\cosh ^{-1} t=\ln (t+\sqrt{t^{2}-1})\) and by considering the cases \(x \geq 0\) and \(x<0.\)

Let $$f(x)=\left\\{\begin{array}{cl}x & \text { if } 0 \leq x \leq 2 \\\2 x-2 & \text { if } 2

Find the \(x\) -coordinate of the point(s) of inflection of \(f(x)=\operatorname{sech} x .\) Report exact answers in terms of logarithms (use Theorem 6.10 ).

Find the volume of the solid generated in the following situations. The region \(R\) in the first quadrant bounded by the graphs of \(y=x\) and \(y=1+\frac{x}{2}\) is revolved about the line \(y=3\).
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