Problem 50 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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 1: 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: $$\frac{du}{dx} = \frac{-2}{x^2}$$

Step 2: Find the derivative of $$\operatorname{csch}^{-1}$$ with respect to $$u$$

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

Step 3: 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: $$\frac{df}{dx} = -\frac{1}{(\frac{2}{x})\sqrt{1+(\frac{2}{x})^2}} \cdot \frac{-2}{x^2}$$

Step 4: 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 $$\frac{df}{dx} = \frac{x}{\sqrt{x^2+4}}$$.

Key Concepts

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.