Problem 24 - Thomas Calculus

Question

In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\left(4 x^{2}-1\right) \operatorname{csch}(\ln 2 x)$$

Step-by-Step Solution

Answer

The derivative is \( y' = 8x \operatorname{csch}(\ln 2x) - \frac{4x^2 - 1}{x} \operatorname{csch}(\ln 2x) \coth(\ln 2x) \).

Step 1: Identify the Functions

The given function to differentiate is \( y= (4x^2-1) \cdot \operatorname{csch}(\ln 2x) \). This is a product of two functions: \( u(x) = 4x^2 - 1 \) and \( v(x) = \operatorname{csch}(\ln 2x) \).

Step 2: Apply the Product Rule

The derivative of a product \( u(x) v(x) \) is given by the product rule: \( (uv)' = u'v + uv' \). Here, we need to calculate \( u' \) and \( v' \).

Step 3: Differentiate \( u(x) \)

Differentiate \( u(x) = 4x^2 - 1 \). Using the power rule, \( u'(x) = 8x \).

Step 4: Differentiate \( v(x) \)

To differentiate \( v(x) = \operatorname{csch}(\ln 2x) \), use the chain rule. The derivative of \( \operatorname{csch}(x) \) is \( -\operatorname{csch}(x)\coth(x) \). So, \( v'(x) = -\operatorname{csch}(\ln 2x) \cdot \coth(\ln 2x) \cdot \frac{1}{x} \).

Step 5: Compute \( y' \) Using the Product Rule

Now substitute back into the product rule: \( y' = (8x)(\operatorname{csch}(\ln 2x)) + (4x^2 - 1)(-\operatorname{csch}(\ln 2x) \cdot \coth(\ln 2x) \cdot \frac{1}{x}) \).

Step 6: Simplify the Expression

Carefully simplify the expression for \( y' \): \[ y' = 8x \operatorname{csch}(\ln 2x) - \frac{4x^2 - 1}{x} \operatorname{csch}(\ln 2x) \coth(\ln 2x) \]. Further simplification would depend on any specific requirements or forms needed.

Key Concepts

Product Rule

The Product Rule is a crucial tool in calculus for finding the derivative of a product of two functions. If you have a function like the one in this exercise, where your function is expressed as the product of two separate functions, it is the product rule that simplifies this calculation.

• Formula: The derivative of a function that is the product of two functions, say \( u(x) \) and \( v(x) \), is expressed as \( (uv)' = u'v + uv' \).
• Application: First, identify the functions you are multiplying. In our exercise, we have \( u(x) = 4x^2 - 1 \) and \( v(x) = \operatorname{csch}(\ln 2x) \).
• Calculate each derivative separately: Find \( u'(x) \) by differentiating \( 4x^2 - 1 \), and use another rule like Chain Rule to find \( v'(x) \).

With these pieces, you can apply the product rule to find the overall derivative by adding the products of these derivatives together. This multi-step process makes handling complex products straightforward.

Chain Rule

The Chain Rule is indispensable when differentiating compositions of functions. It is especially used when you have a function inside another function.

• Understanding Composition: Suppose you have a function \( g(f(x)) \). The outer function is \( g \) and the inner function is \( f \).
• Formula: The derivative using the chain rule is \( \frac{d}{dx}g(f(x)) = g'(f(x)) \cdot f'(x) \).
• Application: Let's apply this to our function \( v(x) = \operatorname{csch}(\ln 2x) \). The function \( g(x) = \operatorname{csch}(x) \) and the inner function \( f(x) = \ln 2x \).

To differentiate \( \operatorname{csch}(\ln 2x) \), apply the chain rule to find the derivative, since the derivative of \( \operatorname{csch}(x) \) is \(-\operatorname{csch}(x)\coth(x)\) and the derivative of \( \ln 2x \) is \( \frac{1}{x} \). Together, these derivatives give you the complete derivative of the composite function. This rule helps us manage these nested functions efficiently.

Hyperbolic Functions

Hyperbolic functions, like \( \operatorname{csch}(x) \), are analogous to the trigonometric functions but relate to hyperbolas rather than circles.

• Definition: The hyperbolic cosecant \( \operatorname{csch}(x) \) is defined as \( \frac{1}{\sinh(x)} \), where \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).
• Important Derivatives: The derivative of \( \operatorname{csch}(x) \) is \(-\operatorname{csch}(x)\coth(x)\), where \( \operatorname{coth}(x) = \frac{\cosh(x)}{\sinh(x)} \).
• Application: In the exercise, the function involved the derivative of \( \operatorname{csch}(\ln 2x) \), requiring an understanding of both the hyperbolic and natural logarithm functions together.

Hyperbolic functions are used in various fields such as physics and engineering, especially when working with wave equations or areas involving hyperbolic geometry. Understanding how to differentiate them broadens your ability to tackle many mathematical problems.