Problem 22 - Thomas Calculus

Question

In Exercises \(13-24,\) 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 \(\frac{dy}{dv} = \coth v - \coth v \times \text{csch}^2 v\).

Step 1 - Differentiate the Natural Logarithm Term

To differentiate the function \(y = \ln \sinh v\), apply the chain rule. The derivative of \(\ln u\) is \(\frac{1}{u}\), so for \(\ln \sinh v\), the derivative is \(\frac{1}{\sinh v}\). Next, differentiate \(\sinh v\) with respect to \(v\), which is \(\cosh v\). Combine these using the chain rule: \(\frac{1}{\sinh v} \times \cosh v = \coth v\).

Step 2 - Differentiate the Hyperbolic Cotangent Term

Given part of the equation \(-\frac{1}{2} \coth^2 v\), use the chain rule to differentiate. The derivative of \(u^n\) is \(n \times u^{n-1} \times \frac{du}{dv}\). For \(\coth^2 v\), \(u^n = \coth^2 v\) and \(n = 2\), giving \(-1\times \coth v \times (-\text{csch}^2 v) = \coth v \times \text{csch}^2 v\). Simplifying, we get \(-\coth v \times \text{csch}^2 v\).

Step 3 - Combine the Derivatives

Add the derivatives obtained from Step 1 and Step 2. The overall derivative \(\frac{dy}{dv}\) is \(\coth v - \coth v \times \text{csch}^2 v\).

Step 4 - Simplify the Expression

Simplify the derivative expression if possible. Since both terms share a common factor \(\coth v\), factor it out if needed. The fully simplified derivative is \(\frac{dy}{dv} = \coth v - \coth v \times \text{csch}^2 v\).

Key Concepts

Chain Rule

The chain rule is a powerful tool in differential calculus that allows us to differentiate composite functions. It's essential when you're dealing with functions nested inside each other. Imagine you have a function of a function, like in our exercise where \( y = \ln \sinh v\). Here, \( \sinh v\) is inside the logarithmic function. The chain rule helps by breaking the differentiation into manageable steps. \Firstly, you differentiate the outer function and then multiply it by the derivative of the inner function. For \(\ln \sinh v\), the outer function is \(\ln u\), with the derivative \(\frac{1}{u}\). You link it to the inner function, \(\sinh v\), which has the derivative \(\cosh v\). Therefore, by applying the chain rule, you multiply these derivatives: \(\frac{1}{\sinh v} \times \cosh v \), which simplifies to \(\coth v\).

• Differentiate the outer function first.
• Then, differentiate the inner function.
• Multiply both derivatives for the final result.

Understanding and mastering the chain rule is crucial because it opens up many possibilities in calculus. It allows you to differentiate complex structures by breaking them down into simpler parts.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for the hyperbola, much like sine and cosine relate to the circle. The main hyperbolic functions are \(\sinh x\) for hyperbolic sine and \(\cosh x\) for hyperbolic cosine. In our exercise, \(\sinh v\) is used inside a logarithm, while \(-\frac{1}{2} \coth^2 v\) involves the hyperbolic cotangent \(\coth v\).
Hyperbolic functions possess unique derivatives. For example, the derivative of \(\sinh v\) is \(\cosh v\), and the derivative of \(\coth v\) is \(-\text{csch}^2 v\). An interesting feature is how these functions interrelate:

• \(\sinh^2 v + \cosh^2 v = \cosh^2 v - \sinh^2 v = 1\).
• \(\frac{d}{dv}\sinh v = \cosh v\).
• \(\frac{d}{dv}\coth v = -\text{csch}^2 v\).

This knowledge extends your understanding of calculus, especially as hyperbolic functions are often used in solving differential equations and modeling various phenomena, from physics to engineering.

Derivative Simplification

Derivative simplification is a crucial step in making your solution understandbale and concise. After you've calculated the derivatives of each part of the function, you often find areas where you can combine like terms or factor out common elements. Let's take our example further.

In Step 3, the derivatives you found were \(\coth v\) from the first term, and \(-\coth v \times \text{csch}^2 v\) from the second term. Notice the common factor \(\coth v\). You can factor it out to simplify the expression.

• Identify common factors in the terms.
• Factor them out to simplify.
• Simplified forms are not only easier to work with but also help in understanding and communicating solutions.

The final expression becomes \(\coth v (1 - \text{csch}^2 v)\), which shows how simplification leads to a clearer, more concise derivative. Simplifying derivatives is essential not only in terms of mathematical aesthetics but also for practical applications, making further operations more straightforward and less error-prone.