Problem 20 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The derivative is \(-\operatorname{csch}(\theta)\operatorname{coth}(\theta)(1-\ln(\operatorname{csch}(\theta))) + \operatorname{csch}(\theta)\coth(\theta)\)."

Step 1: Identify the expression

The expression we need to differentiate is given as: \[ y = \operatorname{csch}(\theta)(1-\ln(\operatorname{csch}(\theta))) \]This involves the hyperbolic cosecant function and a logarithmic term.

Step 2: Apply the Product Rule

The expression is a product of two functions: \( f(\theta) = \operatorname{csch}(\theta) \) and \( g(\theta) = 1 - \ln(\operatorname{csch}(\theta)) \). Apply the product rule: \[ \frac{d}{d\theta}[f(\theta)g(\theta)] = f'(\theta)g(\theta) + f(\theta)g'(\theta) \].

Step 3: Differentiate each component

Differentiate \( f(\theta) = \operatorname{csch}(\theta) \): \[ f'(\theta) = -\operatorname{csch}(\theta)\operatorname{coth}(\theta) \].Differentiate \( g(\theta) = 1 - \ln(\operatorname{csch}(\theta)) \): \[ g'(\theta) = \frac{d}{d\theta}[-\ln(\operatorname{csch}(\theta))] = \coth(\theta) \].

Step 4: Substitute derivatives in product rule

Using the derivatives from Step 3, substitute them into the product rule equation from Step 2:\[ \frac{d}{d\theta}[y] = (-\operatorname{csch}(\theta)\operatorname{coth}(\theta))(1-\ln(\operatorname{csch}(\theta))) + \operatorname{csch}(\theta)\cdot \coth(\theta) \].

Step 5: Simplify the expression

Simplify the expression obtained in Step 4:\[ \frac{d}{d\theta}[y] = -\operatorname{csch}(\theta)\operatorname{coth}(\theta)(1-\ln(\operatorname{csch}(\theta))) + \operatorname{csch}(\theta)\coth(\theta) \].This can be further simplified by factoring out common terms, but the expression is correct at this stage.

Key Concepts

Product Rule

In calculus, the product rule is a technique used to differentiate expressions where two functions are multiplied together. This is essential because the derivative of a product is not simply the product of the derivatives. Instead, the product rule states that if you have two functions, say \( f(x) \) and \( g(x) \), their derivative \( (f(x)g(x))' \) is

• \( f'(x)g(x) + f(x)g'(x) \).

Applying the product rule correctly requires you to:

• Identify the two individual functions involved in the product.
• Differentiate each function separately with respect to the variable.
• Substitute these derivatives into the product rule formula.

For the expression \( y = \operatorname{csch}(\theta)(1-\ln(\operatorname{csch}(\theta))) \), we identified:

• \( f(\theta) = \operatorname{csch}(\theta) \)
• \( g(\theta) = 1 - \ln(\operatorname{csch}(\theta)) \)

By using the product rule, we successfully differentiated each part to find the derivative of \( y \). This method ensures we account for how both parts of the product change with respect to the variable \( \theta \).
Understanding the product rule is crucial for dealing with more complex calculus problems where simple differentiation strategies don't suffice.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola, rather than a circle. Common hyperbolic functions include \( \sinh \), \( \cosh \), and \( \operatorname{csch} \), which stand for hyperbolic sine, cosine, and cosecant, respectively. The function \( \operatorname{csch}(\theta) \) is defined as:

• \( \operatorname{csch}(\theta) = \frac{1}{\sinh(\theta)} \)

This means it is the reciprocal of the hyperbolic sine function, \( \sinh(\theta) \).
When differentiating hyperbolic functions, it’s vital to know their basic derivatives. For instance:

• Derivative of \( \operatorname{csch}(\theta) \) is \( -\operatorname{csch}(\theta)\operatorname{coth}(\theta) \)

During the differentiation process, these derivatives play a critical role. In our exercise, we used the derivative of \( \operatorname{csch}(\theta) \) to apply the product rule correctly.
Understanding hyperbolic functions and their properties significantly broadens your ability to handle diverse mathematical problems, especially those involving exponential growth or hyperbolic geometry.

Logarithmic Differentiation

Logarithmic differentiation is a powerful tool for finding derivatives, especially when dealing with products, quotients, or complex functions involving logarithms. The main concept is to take the natural logarithm of both sides of an equation and then differentiate implicitly. This technique is particularly useful when the function involves terms nested within a logarithm.
Consider the function \( g(\theta) = 1 - \ln(\operatorname{csch}(\theta)) \). Here, the differentiation inside reveals useful properties. To differentiate \( -\ln(\operatorname{csch}(\theta)) \), you use:

• \( \frac{d}{d\theta}[-\ln(\operatorname{csch}(\theta))] = \coth(\theta) \)

This makes the expression manageable and clarifies how logarithmic functions behave when differentiated.
Logarithmic differentiation is an essential method in calculus because it simplifies the differentiation process of functions that seem cumbersome or intricate at first glance. Mastery of this approach widens the array of problems you can solve efficiently.

Derivative Simplification

Once you've found the derivative using rules like the product rule, simplification becomes important. Simplifying derivatives makes them easier to interpret and can reveal important features about the behavior of the function. After the product rule and differentiation steps, we were left with:\[ \frac{d}{d\theta}[y] = -\operatorname{csch}(\theta)\operatorname{coth}(\theta)(1-\ln(\operatorname{csch}(\theta))) + \operatorname{csch}(\theta)\coth(\theta) \]Simplifying involves:

• Combining like terms.
• Factoring out common factors, if possible.
• Reducing the expression to its simplest form.

While the expression from our solution might seem complex, deducing it to a simpler form can greatly aid in understanding the derivative’s implications in real-world contexts.
Derivative simplification is crucial because it reduces potential errors in interpretation and allows easier application of the derivative. Whether for graphing, analysis, or theoretical exploration, a simplified derivative facilitates better mathematical communication and interpretation.