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 \( \frac{dy}{dv} = \coth v + \operatorname{coth} v \cdot \operatorname{csch}^2 v \).

Step 1: Differentiate the First Term

We need to differentiate each part of the function separately. The first term is \( y_1 = \ln (\sinh v) \). The derivative of \( \ln f(x) \) is \( \frac{f'(x)}{f(x)} \). Therefore, we need to find the derivative of \( \sinh v \), which is \( \cosh v \). Thus, the derivative of \( y_1 \) with respect to \( v \) is \( \frac{\cosh v}{\sinh v} = \coth v \).

Step 2: Differentiate the Second Term

The second term is \( y_2 = -\frac{1}{2} \operatorname{coth}^2 v \). To differentiate this, recall that if \( y = u^2 \), then \( \frac{d}{dv}(u^2) = 2u \cdot \frac{du}{dv} \). Here, \( u = \operatorname{coth} v \). The derivative of \( \operatorname{coth} v \) is \( -\operatorname{csch}^2 v \). Therefore, the derivative of \( y_2 \) is \( -\frac{1}{2} \cdot 2 \operatorname{coth} v \cdot (-\operatorname{csch}^2 v) = \operatorname{coth} v \cdot \operatorname{csch}^2 v \).

Step 3: Combine the Derivatives

To find the total derivative of \( y \), add the results from Step 1 and Step 2. The derivative of \( y \) with respect to \( v \) is: \[ \frac{dy}{dv} = \coth v + \operatorname{coth} v \cdot \operatorname{csch}^2 v \]. This is the derivative of the given function.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the trigonometric functions but for hyperbolas instead of circles. Just like trigonometric functions are based on the unit circle, hyperbolic functions are based on a unit hyperbola. The most common hyperbolic functions are \( \sinh(x) \) and \( \cosh(x) \), defined as follows:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

These functions resemble the sine and cosine functions, hence their names, but they differ significantly. They are useful in various applications, such as modeling real-world phenomena that involve hyperbolic geometry.
The derivatives of hyperbolic functions are quite similar to those of trigonometric functions, making them easier to work with. For example, the derivative of \( \sinh(v) \) is \( \cosh(v) \), just as the derivative of \( \sin(x) \) is \( \cos(x) \).
Understanding these functions allows you to solve complex problems in calculus involving hyperbolic terms.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions that are products, quotients, or powers, where applying standard rules would be cumbersome. This method simplifies the process by taking the natural logarithm of both sides of the equation.
For instance, consider a function \( y = f(v) \). By taking the natural logarithm, we have \( \ln(y) = \ln(f(v)) \). Differentiating both sides with respect to \( v \) gives \( \frac{d}{dv}[\ln(y)] = \frac{d}{dv}[\ln(f(v))] \). The derivative of \( \ln(y) \) is \( \frac{1}{y} \frac{dy}{dv} \), and the derivative of \( \ln(f(v)) \) uses the chain rule. This transforms into:

• \( y' = y \frac{f'(v)}{f(v)} \)

This process makes differentiating complex functions more manageable, especially when handling product or quotient forms, and was demonstrated in the differentiation of \( y = \ln(\sinh v) \).

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variable where the functions are defined. These identities are helpful in simplifying expressions and are used extensively in calculus and other areas of mathematics.
In the context of hyperbolic functions, identities similar to trigonometric identities often apply. One important pair is:

• \( \coth(v) = \frac{\cosh(v)}{\sinh(v)} \)
• \( \csch(v) = \frac{1}{\sinh(v)} \)

These identities allow us to express derivatives in simpler forms, as seen in the differentiation of \( y = \ln(\sinh v) - \frac{1}{2}\operatorname{coth}^2 v \). Utilizing these identities can streamline the process by reducing complex functions into more solvable components.

Calculus Exercises

Calculus exercises are essential for mastering the principles and applications of calculus. They provide opportunities to apply concepts like derivatives, limits, integration, and more to solve mathematical problems.
One common calculus exercise involves finding the derivative of a function. This might include employing techniques such as the power rule, product rule, quotient rule, and chain rule. In more complex situations, logarithmic differentiation and the use of identities come into play, as they simplify the process for intricate functions.
When practicing calculus exercises, ensure you understand each step and the reason behind it. This deep understanding will equip you with problem-solving skills necessary for more advanced topics.