Problem 50 - University Calculus: Early Transcendentals

Question

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\frac{\theta \sin \theta}{\sqrt{\sec \theta}}$$

Step-by-Step Solution

Answer

\( \frac{dy}{d\theta} = \frac{\theta \sin \theta}{\sqrt{\sec \theta}} \left( \frac{1}{\theta} + \cot \theta - \frac{1}{2} \tan \theta \right) \)

Step 1: Identify the components of the function

Given the function \( y = \frac{\theta \sin \theta}{\sqrt{\sec \theta}} \), identify the numerator and denominator. The numerator is \( \theta \sin \theta \) and the denominator is \( \sqrt{\sec \theta} \).

Step 2: Apply logarithmic differentiation

Take the natural logarithm of both sides of the equation: \( \ln y = \ln \left( \frac{\theta \sin \theta}{\sqrt{\sec \theta}} \right) \). Expand the expression using properties of logarithms: \( \ln y = \ln (\theta) + \ln (\sin \theta) - \frac{1}{2} \ln (\sec \theta) \).

Step 3: Differentiate both sides with respect to \( \theta \)

Differentiate \( \ln y \) using the chain rule: \( \frac{1}{y} \frac{dy}{d\theta} \). For the right side, differentiate term by term: \( \frac{d}{d\theta}(\ln \theta) = \frac{1}{\theta} \), \( \frac{d}{d\theta}(\ln \sin \theta) = \cot \theta \), and \( \frac{d}{d\theta} \left( -\frac{1}{2} \ln \sec \theta \right) = -\frac{1}{2} \cdot \tan \theta \). This gives the equation: \( \frac{1}{y} \frac{dy}{d\theta} = \frac{1}{\theta} + \cot \theta - \frac{1}{2} \tan \theta \).

Step 4: Solve for the derivative \( \frac{dy}{d\theta} \)

Multiply both sides of the equation by \( y \) to solve for \( \frac{dy}{d\theta} \): \( \frac{dy}{d\theta} = y \left( \frac{1}{\theta} + \cot \theta - \frac{1}{2} \tan \theta \right) \). Substitute back \( y = \frac{\theta \sin \theta}{\sqrt{\sec \theta}} \): \( \frac{dy}{d\theta} = \frac{\theta \sin \theta}{\sqrt{\sec \theta}} \left( \frac{1}{\theta} + \cot \theta - \frac{1}{2} \tan \theta \right) \). Simplify to obtain the final expression for the derivative.

Key Concepts

Derivative Calculation

Calculating derivatives is a fundamental concept in calculus. It involves finding how a function changes at any given point. A derivative provides the rate at which a function is changing. For the given function \( y = \frac{\theta \sin \theta}{\sqrt{\sec \theta}} \), we will use logarithmic differentiation to find the derivative.
This can be especially helpful when dealing with complex expressions, as it allows breaking down the problem into simpler terms.
In our exercise, after taking the natural log of both sides:

• We simplify the expression using laws of logarithms, which help convert products into sums and quotients into subtractions.
• Then, we differentiate each term separately. This makes it easier to handle the complexity of the original function.

The derivative is computed in the context of \( \theta \), indicating how \( y \) changes as \( \theta \) increases or decreases.

Trigonometric Functions

Trigonometric functions play a significant role in calculus problems, including the one given here. In this function, both \( \sin \theta \) and \( \sec \theta \) are trigonometric functions involved in the problem.
Understanding the derivatives of these functions is crucial:

• The derivative of \( \sin \theta \) with respect to \( \theta \) is \( \cos \theta \).
• \( \sec \theta \) is the reciprocal of \( \cos \theta \), hence its derivative is related to the tangent function, specifically \( \sec \theta \tan \theta \).

The problem also uses the property that \( \sec \theta = \frac{1}{\cos \theta} \), leading to \( \ln(\sec \theta) = -\ln(\cos \theta) \). These relations are vital in simplifying complex expressions involving trigonometric components during differentiation.

Chain Rule in Calculus

The chain rule is a powerful tool in calculus used to differentiate composite functions. It states that to differentiate a composition of functions, you must multiply the derivative of the outer function by the derivative of the inner function.
This rule is crucial for calculating the derivative in our logarithmic differentiation problem, as it involves a combination of functions:

• When differentiating \( \ln y \), it requires using the chain rule to find \( \frac{1}{y} \cdot \frac{dy}{d\theta} \).
• Each logarithm component, like \( \ln(\sin \theta) \) or \( -\frac{1}{2} \ln(\sec \theta) \), is a function of \( \theta \), and requires chaining its differentiation with respect to \( \theta \).

The chain rule allows us to systematically handle the differentiation of products or quotients of functions, essential for obtaining a clear and accurate derivative result.