Problem 34 - Thomas Calculus

Question

In Exercises \(25-36,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\operatorname{csch}^{-1} 2^{\theta} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{-\ln(2)}{\sqrt{4^\theta + 1}} \).

Step 1: Identify the Inverse Hyperbolic Function

The given function is in the form \( y = \operatorname{csch}^{-1}(u) \) where \( u = 2^{\theta} \). The derivative of the inverse hyperbolic cosecant function, \( \operatorname{csch}^{-1}(u) \), with respect to \( u \) is \( \frac{-1}{|u|\sqrt{u^2+1}} \).

Step 2: Differentiate the Inner Function

We need to find the derivative of the inner function \( u = 2^{\theta} \) with respect to \( \theta \). Using the chain rule, derive \( \frac{d}{d\theta}(2^{\theta}) = 2^{\theta} \ln(2) \).

Step 3: Apply the Chain Rule

Using the chain rule, the derivative of \( y \) with respect to \( \theta \) is \( \frac{dy}{d\theta} = \frac{dy}{du} \cdot \frac{du}{d\theta} \). Substitute the derivatives from Step 1 and Step 2: \( \frac{dy}{d\theta} = \frac{-1}{|2^\theta| \sqrt{(2^\theta)^2+1}} \cdot 2^\theta \ln(2) \).

Step 4: Simplify the Expression

Simplify the expression derived from the chain rule application: \( \frac{dy}{d\theta} = \frac{-2^{\theta} \ln(2)}{|2^\theta| \sqrt{4^\theta + 1}} \). Since \( 2^\theta \) is always positive, \( |2^\theta| = 2^\theta \). Therefore, \( \frac{dy}{d\theta} = \frac{-\ln(2)}{\sqrt{4^\theta + 1}} \).

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions are the inverses of the hyperbolic functions like sinh, cosh, and tanh. The function we are focusing on here is the inverse hyperbolic cosecant, denoted as \( \operatorname{csch}^{-1}(x) \). It is the inverse of the hyperbolic cosecant function, which itself is defined as \( \operatorname{csch}(x) = \frac{1}{\sinh(x)} \).

Hyperbolic functions resemble their trigonometric counterparts but are based on hyperbolas instead of circles. They have a unique set of properties and are used in various calculus and mathematical applications.

• Definition: The inverse hyperbolic cosecant of \( u \) can be expressed in terms of logarithms: \( \operatorname{csch}^{-1}(u) = \ln \left(\frac{1}{u} + \sqrt{\frac{1}{u^2} + 1}\right) \), for \( u > 0 \).
• Derivative: The derivative of \( \operatorname{csch}^{-1}(u) \) with respect to \( u \) is \( \frac{-1}{|u|\sqrt{u^2 + 1}} \).

Chain Rule

The chain rule is a fundamental differentiation technique used to find the derivative of composite functions. A composite function is a function that is made up of two or more functions nested within each other.

In the exercise, we have a composite function \( y = \operatorname{csch}^{-1}(2^{\theta}) \). The chain rule allows us to differentiate \( y \) with respect to \( \theta \) by breaking it down into simpler parts.

• Apply the Chain Rule: If you have a composite function \( y = f(g(x)) \), the derivative is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
• Inner and Outer Functions: In our specific problem, the outer function is \( \operatorname{csch}^{-1}(u) \) and the inner function is \( u = 2^{\theta} \). We differentiate the outer function with respect to the inner function, and then the inner function with respect to \( \theta \).

Differentiation Techniques

Differentiation techniques allow us to find the rate at which a function is changing. In calculus, these techniques are essential for solving problems involving rates of change and slopes of curves.

In this exercise, we utilized several differentiation techniques:

• Recognize the Function: Identify the type of function you are dealing with (e.g., exponential, trigonometric, or in this case, hyperbolic).
• Inverse Function Derivative: Use known derivative rules for inverse functions, like \( \operatorname{csch}^{-1}(x) \).
• Logarithmic Differentiation: When differentiating exponential functions, logarithmic properties can be helpful. For instance, the derivative of \( 2^{\theta} \) is \( 2^{\theta} \ln(2) \), using the formula \( \frac{d}{dx}(a^x) = a^x \ln(a) \).
• Simplification: After applying differentiation rules, always simplify your expression, as seen when we simplified \( \frac{dy}{d\theta} \) to \( \frac{-\ln(2)}{\sqrt{4^\theta + 1}} \).

Understanding these techniques makes it easier to approach and solve complex calculus problems, enhancing problem-solving skills.