Problem 32 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=\ln \left(\frac{\sqrt{\sin \theta \cos \theta}}{1+2 \ln \theta}\right)$$

Step-by-Step Solution

Answer

The derivative is \( y' = \frac{1}{2} \cot \theta - \frac{1}{2} \tan \theta - \frac{2}{\theta(1 + 2\ln \theta)} \).

Step 1: Simplify the Expression

First, let's rewrite the function \( y = \ln \left( \frac{\sqrt{\sin \theta \cos \theta}}{1+2 \ln \theta} \right) \) for simpler differentiation. Use the logarithm property: \( \ln \left( \frac{a}{b} \right) = \ln a - \ln b \). This gives us:\[ y = \ln \left( \sqrt{\sin \theta \cos \theta} \right) - \ln \left( 1 + 2 \ln \theta \right) \]The next step is to simplify further by addressing each term separately.

Step 2: Simplify Each Term

For the first term, use the property \( \ln(\sqrt{u}) = \frac{1}{2} \ln(u) \):\[ \ln \left( \sqrt{\sin \theta \cos \theta} \right) = \frac{1}{2} (\ln(\sin \theta) + \ln(\cos \theta)) \]This simplifies the function to:\[ y = \frac{1}{2}(\ln(\sin \theta) + \ln(\cos \theta)) - \ln(1 + 2 \ln \theta) \]

Step 3: Differentiate the Simplified Expression

Now, take the derivative of each part with respect to \( \theta \). - Differentiate \(\frac{1}{2}(\ln(\sin \theta) + \ln(\cos \theta))\): \[ \frac{d}{d\theta}\frac{1}{2}(\ln(\sin \theta)) = \frac{1}{2} \frac{1}{\sin \theta} \cdot \cos \theta = \frac{1}{2} \cot \theta \] \[ \frac{d}{d\theta}\frac{1}{2}(\ln(\cos \theta)) = \frac{1}{2} \frac{1}{\cos \theta} \cdot (-\sin \theta) = -\frac{1}{2} \tan \theta \]- Differentiate \(-\ln(1 + 2 \ln \theta)\) using the chain rule: \[ \frac{d}{d\theta}(-\ln(1 + 2 \ln \theta)) = -\frac{1}{1 + 2\ln \theta} \cdot \frac{d}{d\theta}(1 + 2\ln \theta) = -\frac{2}{(1 + 2\ln \theta)\theta} \]Combine all derivatives:\[ y' = \frac{1}{2} \cot \theta - \frac{1}{2} \tan \theta - \frac{2}{\theta(1 + 2\ln \theta)} \]

Step 4: Write the Final Derivative Expression

The derivative of \( y \) with respect to \( \theta \) is therefore:\[ y' = \frac{1}{2} \cot \theta - \frac{1}{2} \tan \theta - \frac{2}{\theta(1 + 2\ln \theta)} \]

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a technique used to simplify the process of differentiating complex functions, especially those involving products, quotients, or powers that can be cumbersome. This method utilizes the natural logarithm to transform products or quotients into a sum or difference, making them easier to differentiate.

When faced with a function that's a quotient or has roots, like in the given exercise, we start by taking the natural logarithm of both sides. The properties of logarithms, such as the logarithm of a quotient, become crucial here. Specifically, \[ \ln \left( \frac{a}{b} \right) = \ln a - \ln b \],which breaks down the function into simpler additive components.

Once the logarithm is applied, differentiation becomes more manageable. The derivative of \( \ln u \) with respect to \( u \) is \( \frac{1}{u} \left( \frac{du}{dx} \right) \), allowing us to apply other rules, such as the chain rule, in subsequent steps. This makes logarithmic differentiation a powerful tool, especially when encountering complex expressions in calculus.

Trigonometric Functions

Trigonometric functions like sine (\( \sin \theta \)) and cosine (\( \cos \theta \)) often appear in calculus problems. These functions are periodic and have specific properties that simplify calculations.

In our exercise, both \( \sin \theta \) and \( \cos \theta \) are present, and the logarithm of their product is taken. Using the properties of logarithms, \( \ln (ab) = \ln a + \ln b \), we simplify \( \ln (\sin \theta \cos \theta) \) to \( \ln (\sin \theta) + \ln (\cos \theta) \).

Differentiating trigonometric functions involves remembering their respective derivatives:

• \( \frac{d}{d\theta} \sin \theta = \cos \theta \)
• \( \frac{d}{d\theta} \cos \theta = -\sin \theta \)
• \( \cot \theta = \frac{\cos \theta}{\sin \theta} \)
• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)

In the solution, these derivatives contribute to the terms \( \frac{1}{2} \cot \theta \) and \( -\frac{1}{2} \tan \theta \), emphasizing the role of trigonometric identities in differentiation.

Chain Rule

The chain rule is a fundamental tool in calculus for differentiating composite functions, where one function is nested inside another. This rule is essential when dealing with expressions like \( -\ln(1 + 2\ln \theta) \).

The general form of the chain rule states: If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). Essentially, we differentiate the outer function and multiply it by the derivative of the inner function.

In the exercise, after applying the logarithmic differentiation, the chain rule is used to differentiate \( \ln(1 + 2\ln \theta) \). This step involves:

• Differentiating the outer function: \( \frac{-1}{1 + 2\ln \theta} \)
• Differentiating the inner function: \( 2\left( \frac{1}{\theta} \right) \)

Thus, the combination yields \( -\frac{2}{\theta(1 + 2\ln \theta)} \), demonstrating how the chain rule intertwines functions for accurate differentiation.

Logarithmic Properties

Logarithmic properties offer invaluable simplifications in calculus. In dealing with complex functions, recognizing how to transform expressions using logarithms can save time and effort.

Key properties utilized in the exercise include:

• The quotient rule for logs: \( \ln \left( \frac{a}{b} \right) = \ln a - \ln b \)
• The power rule for logs: \( \ln (a^n) = n \ln a \)
• The property for the square root: \( \ln(\sqrt{u}) = \frac{1}{2} \ln(u) \)

For example, in the given function, by applying \( \ln(\sqrt{\sin \theta \cos \theta}) = \frac{1}{2}(\ln(\sin \theta) + \ln(\cos \theta)) \), we break down potentially unwieldy expressions into additive forms, allowing for simpler differentiation.

Mastering these properties not only makes differentiation more accessible but also enhances overall mathematical agility in problem-solving contexts, reflecting the importance of logarithmic properties in practical calculus applications.