Problem 19 - Thomas Calculus

Question

In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\operatorname{sech} \theta(1-\ln \operatorname{sech} \theta) $$

Step-by-Step Solution

Answer

The derivative is \(-\operatorname{sech} \theta \cdot \operatorname{tanh} \theta \cdot \ln(\operatorname{sech} \theta)\)."

Step 1: Understand the problem

We need to find the derivative of the given function with respect to \( \theta \). The function is \( y = \operatorname{sech} \theta \cdot (1 - \ln(\operatorname{sech} \theta)) \).

Step 2: Identify the rule to be used

Since the given function is a product of two functions: \( u(\theta) = \operatorname{sech} \theta \) and \( v(\theta) = 1 - \ln(\operatorname{sech} \theta) \), we will use the product rule, which is \( (uv)' = u'v + uv' \).

Step 3: Differentiate each function

First, find the derivative of \( u(\theta) = \operatorname{sech} \theta \): \( u'(\theta) = -\operatorname{sech} \theta \cdot \operatorname{tanh} \theta \).Next, differentiate \( v(\theta) = 1 - \ln(\operatorname{sech} \theta) \):\( v'(\theta) = -\frac{d}{d\theta}[\ln(\operatorname{sech} \theta)] = \operatorname{tanh} \theta \).

Step 4: Apply the product rule

Using the derivatives calculated:\[ y'(\theta) = u'(\theta)v + u(\theta)v' \]Substitute the values:\[ y'(\theta) = (-\operatorname{sech} \theta \cdot \operatorname{tanh} \theta)(1 - \ln(\operatorname{sech} \theta)) + (\operatorname{sech} \theta)(\operatorname{tanh} \theta) \]

Step 5: Simplify the expression

Simplify the expression for \( y'(\theta) \):\[ y'(\theta) = -\operatorname{sech} \theta \cdot \operatorname{tanh} \theta \cdot (1 - \ln(\operatorname{sech} \theta)) + \operatorname{sech} \theta \cdot \operatorname{tanh} \theta \]Combine like terms:\[ y'(\theta) = \operatorname{sech} \theta \cdot \operatorname{tanh} \theta \cdot (1 - \ln(\operatorname{sech} \theta) - 1) \]Simplify further:\[ y'(\theta) = -\operatorname{sech} \theta \cdot \operatorname{tanh} \theta \cdot \ln(\operatorname{sech} \theta) \]

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used to find the derivative of a product of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), that are multiplied together, and you need to find the derivative of their product, you can't simply multiply their derivatives. This is where the product rule comes in handy.

According to the product rule, the derivative of the product of two functions \( u(x) \) and \( v(x) \) is given by:

• \( (uv)' = u'v + uv' \)

This means you take the derivative of the first function \( u(x) \) (which is \( u' \)) and multiply it by the second function \( v(x) \), then add the product of the first function \( u(x) \) and the derivative of the second function \( v' \).

For example, in the problem, we have \( u(\theta) = \operatorname{sech} \theta \) and \( v(\theta) = 1 - \ln(\operatorname{sech} \theta) \). Using the product rule, the derivative \( y'(\theta) \) involves differentiating each part independently and combining them as per the rule. This method ensures that the properties of both functions are correctly taken into account in the differentiation process.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola. These functions are important in many areas of calculus and applied mathematics. The basic hyperbolic functions include the hyperbolic sine \( \sinh \), the hyperbolic cosine \( \cosh \), and the hyperbolic tangent \( \tanh \).

In this exercise, we specifically use the hyperbolic secant function, \( \operatorname{sech}(\theta) \), and the hyperbolic tangent function, \( \operatorname{tanh}(\theta) \). The function \( \operatorname{sech}(\theta) \) is defined as:

• \( \operatorname{sech}(\theta) = \frac{1}{\cosh(\theta)} \)

The hyperbolic tangent \( \operatorname{tanh}(\theta) \) can be expressed as:

• \( \operatorname{tanh}(\theta) = \frac{\sinh(\theta)}{\cosh(\theta)} \)

These functions have derivatives that are just as unique as their definitions. For instance, if you find the derivative of \( \operatorname{sech}(\theta) \), you get \(-\operatorname{sech}(\theta) \cdot \operatorname{tanh}(\theta) \). Likewise, the derivative of \( \operatorname{tanh}(\theta) \) is \( \operatorname{sech}^2(\theta) \).

These derivative properties are utilized in calculus exercises to tackle more complex differentiation involving hyperbolic functions.

Calculus Exercises

Calculus exercises often involve finding derivatives of functions involving basic rules such as the product rule, quotient rule, and chain rule. Each of these exercises is an opportunity to practice applying these fundamental concepts to solve real-world problems.

When working through calculus exercises, it's important to:

• Identify the functions involved and determine which differentiation rules apply.
• Calculate derivatives methodically, breaking down each component of the given function.
• Simplify the result to the simplest form possible, combining like terms and reducing fractions where necessary.

In exercises like the one provided, you start by recognizing it as a product of two functions, requiring the application of the product rule. The complexity of calculus exercises comes from dealing with various functions and applying the appropriate rules. Each step builds upon the understanding of simpler concepts to tackle more advanced problems in calculus.

Keeping a systematic approach helps avoid errors and allows for a deeper understanding of how mathematical rules come together to describe changes in functions. Therefore, practice and familiarity with these rules enhance problem-solving skills in calculus.