Problem 22 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\ln \sinh v-\frac{1}{2} \operatorname{coth}^{2} v$$

Step-by-Step Solution

Answer

The derivative is \( \operatorname{coth} v (1 + \operatorname{csch}^2 v) \).

Step 1: Understand the Problem

The problem asks us to find the derivative of the function \( y = \ln \sinh v - \frac{1}{2} \operatorname{coth}^{2} v \) with respect to \( v \). This means we need to apply rules of differentiation to the given function.

Step 2: Differentiate the Logarithmic Function

For the term \( \ln \sinh v \), use the chain rule for differentiation. The derivative of \( \ln u \) with respect to \( u \) is \( \frac{1}{u} \), and here \( u = \sinh v \). Thus, we need to further differentiate \( \sinh v \), which gives \( \cosh v \). Therefore, the derivative of \( \ln \sinh v \) is \( \frac{1}{\sinh v} \times \cosh v = \operatorname{coth} v \).

Step 3: Differentiate the Hyperbolic Cotangent Squared Function

For the term \( -\frac{1}{2} \operatorname{coth}^{2} v \), use the chain rule as well. The derivative of \( u^2 \) with respect to \( u \) is \( 2u \). But we multiply by the derivative of \( \operatorname{coth} v \), which is \( -\operatorname{csch}^2 v \). Therefore, the derivative is \(-\frac{1}{2} \times 2 \times \operatorname{coth} v \times (-\operatorname{csch}^2 v) = \operatorname{coth} v \operatorname{csch}^2 v \).

Step 4: Combine the Derivatives

Combine the results from the previous steps: The derivative of the entire function \( y = \ln \sinh v - \frac{1}{2} \operatorname{coth}^{2} v \) is \( \operatorname{coth} v + \operatorname{coth} v \operatorname{csch}^2 v \).

Step 5: Simplify, if Necessary

The expression for the derivative can be written as \( \operatorname{coth} v (1 + \operatorname{csch}^2 v) \), which is already simplified.

Key Concepts

Hyperbolic Functions

Hyperbolic functions, such as sinh (hyperbolic sine) and coth (hyperbolic cotangent), are analogous to trigonometric functions but are defined using exponential functions. These functions are important in various fields, including mathematics and engineering.

• The hyperbolic sine function, denoted as \( \sinh v \), is defined by \( \sinh v = \frac{e^v - e^{-v}}{2} \).
• The hyperbolic cotangent function, \( \operatorname{coth} v \), is given by \( \operatorname{coth} v = \frac{\cosh v}{\sinh v} \). Here, \( \cosh v = \frac{e^v + e^{-v}}{2} \) is the hyperbolic cosine function.
• Key properties include the identities \( \cosh^2 v - \sinh^2 v = 1 \), and derivatives such as \( \frac{d}{dv} \sinh v = \cosh v \) and \( \frac{d}{dv} \operatorname{coth} v = -\operatorname{csch}^2 v \).

These functions showcase unique properties that separate them from regular trigonometric functions, making them powerful tools in the study of hyperbolic geometry and complex analysis.
Understanding these functions’ behaviors, particularly their derivatives, is crucial for solving problems involving hyperbolic functions.

Chain Rule

The chain rule is a fundamental concept in differentiation, especially when dealing with composite functions. It allows us to differentiate functions within functions by breaking them down into simpler parts.In the equation \( y = \ln \sinh v - \frac{1}{2} \operatorname{coth}^2 v \), the chain rule helps differentiate terms where one function is inside another, like \( \ln \sinh v \). Here's how it works:

• For \( \ln \sinh v \), think of it as two functions: the outer function \( \ln(u) \) and the inner function \( u = \sinh v \).
• First, differentiate the outer function: \( \frac{1}{u} \).
• Then, multiply this by the derivative of the inner function: \( \cosh v \), yielding \( \frac{\cosh v}{\sinh v} = \operatorname{coth} v \).

Applying the chain rule ensures that we correctly account for each layer of the function. This rule is essential for managing complex expressions and finding derivatives step by step.

Derivative Simplification

Once you've found the derivative of each component in a function, simplification often makes the expression more manageable. It's about recombining terms and applying algebraic identities and properties.For the function \( y = \ln \sinh v - \frac{1}{2} \operatorname{coth}^2 v \), after differentiating, we obtain:

• The derivative \( \operatorname{coth} v + \operatorname{coth} v \operatorname{csch}^2 v \). Here, \( \operatorname{csch} v \) is the hyperbolic cosecant function, \( \operatorname{csch} v = \frac{1}{\sinh v} \).
• By factoring out \( \operatorname{coth} v \), simplify the expression to \( \operatorname{coth} v (1 + \operatorname{csch}^2 v) \).

This condensed form is easier to interpret and work with. Simplification is a crucial skill, allowing expressions to be more concise and making further calculations simpler and clearer.