Chapter 7: Problem 27
In Exercises \(7-38,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\theta(\sin (\ln \theta)+\cos (\ln \theta)) $$
Short Answer
Expert verified
The derivative is \( y' = 2\cos(\ln \theta) \).
Step by step solution
01
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.
02
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 \( u' = 1 \).
03
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:\( v' = \frac{\cos(\ln \theta)}{\theta} - \frac{\sin(\ln \theta)}{\theta} \).
04
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:\[y' = 2\cos(\ln \theta).\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
- \((uv)' = u'v + uv'\)
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: \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \)
- 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 \( \frac{1}{\theta} \).
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 \(-\sin(x) \).
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 \( \frac{1}{x} \).