Problem 27 - Thomas Calculus

Question

In Exercises \(7-38,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\theta(\sin (\ln \theta)+\cos (\ln \theta)) $$

Step-by-Step Solution

Answer

The derivative is \( y' = 2\cos(\ln \theta) \).

Step 1: Identify the problem type

We need to find the derivative of the function \( y = \theta(\sin(\ln \theta) + \cos(\ln \theta)) \) with respect to \( \theta \). This requires using both the product rule and the chain rule.

Step 2: Apply the product rule

The product rule states that \((uv)' = u'v + uv'\). Let \( u = \theta \) and \( v = \sin(\ln \theta) + \cos(\ln \theta) \). First, find the derivative of \( u \), which is \( u' = 1 \).

Step 3: Differentiate \( v \) using the chain rule

The function \( v = \sin(\ln \theta) + \cos(\ln \theta) \). Use the chain rule for each component. - For \( \sin(\ln \theta) \): \( \frac{d}{d\theta}(\sin(\ln \theta)) = \cos(\ln \theta) \cdot \frac{d}{d\theta}(\ln \theta) = \frac{\cos(\ln \theta)}{\theta} \).- For \( \cos(\ln \theta) \): \( \frac{d}{d\theta}(\cos(\ln \theta)) = -\sin(\ln \theta) \cdot \frac{d}{d\theta}(\ln \theta) = -\frac{\sin(\ln \theta)}{\theta} \).Combine these derivatives:\( v' = \frac{\cos(\ln \theta)}{\theta} - \frac{\sin(\ln \theta)}{\theta} \).

Step 4: Combine results to find \( y' \)

Using the product rule:\[ y' = u'v + uv' = 1 \cdot (\sin(\ln \theta) + \cos(\ln \theta)) + \theta \left(\frac{\cos(\ln \theta)}{\theta} - \frac{\sin(\ln \theta)}{\theta}\right).\]This simplifies to:\[y' = \sin(\ln \theta) + \cos(\ln \theta) + \cos(\ln \theta) - \sin(\ln \theta).\]Thus, the derivative is:\[y' = 2\cos(\ln \theta).\]

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used to find derivatives of products of two functions. Imagine you have two functions, say \( u \) and \( v \), that are multiplying each other. The product rule tells us that the derivative of the product \( uv \) is given by:

• \((uv)' = u'v + uv'\)

This means that we differentiate \( u \) while holding \( v \) constant, and then we differentiate \( v \) while holding \( u \) constant, adding the results together. In the exercise, \( u \) is \( \theta \), and \( v \) is \( \sin(\ln \theta) + \cos(\ln \theta) \). Calculating \( u' \) is straightforward since it's simply \( 1 \), given \( \theta \). This part of the product rule helps in organizing the process of differentiation when dealing with product terms in your function.

Chain Rule

The chain rule is an indispensable tool in calculus for finding the derivative of composite functions. When you have a function inside another function, like \( \sin(\ln \theta) \), the chain rule helps you differentiate it effectively. The chain rule can be stated as:

• If you have \( f(g(x)) \), then: \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \)

This means you take the derivative of the outer function and multiply it by the derivative of the inner function. In the problem, you apply it separately on \( \sin(\ln \theta) \) and \( \cos(\ln \theta) \):

• Derivative of \( \sin(\ln \theta) \) is \( \cos(\ln \theta) \) times the derivative of \( \ln \theta \), which is \( \frac{1}{\theta} \).
• For \( \cos(\ln \theta) \), you get \(-\sin(\ln \theta) \) times \( \frac{1}{\theta} \).

By working through both, you construct the derivative of \( v \) using the chain rule effectively.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, frequently appear in calculus problems and have specific differentiation rules. Understanding how they transform when differentiated is key:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \(-\sin(x) \).

These basic rules are crucial when dealing with more complex trigonometric expressions inside another function, as seen in this problem. When there's a function within these trigonometric identities, such as \( \ln \theta \) in \( \sin(\ln \theta) \) and \( \cos(\ln \theta) \), you apply the chain rule to differentiate them. Recognizing this structure will help simplify the differentiation process and ensure accuracy when solving similar problems.

Logarithmic Functions

Logarithmic functions are another common concept in calculus. They often combine with other functions to form complex expressions. In differentiation, the natural logarithm function \( \ln(x) \) has a simple rule:

• The derivative of \( \ln(x) \) is \( \frac{1}{x} \).

This derivative is important when \( \ln \) is part of a larger function. In our problem, it shows how the derivative of \( \ln \theta \) contributes when applying the chain rule. With trigonometric functions wrapped around logarithmic functions, knowing the derivative of \( \ln(x) \) ensures you correctly apply the chain rule by providing a crucial multiplication term. This concept ties closely with trigonometric differentiation, where recognizing \( \ln(\theta) \) as part of \( \sin(\ln \theta) \) or \( \cos(\ln \theta) \) affects your overall derivative calculation strategy.