Problem 25 - Thomas Calculus in SI Units

Question

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 \( \frac{dy}{d\theta} \) is \( 2\cos(\ln \theta) \).

Step 1: Identify the Components

The given function is \( y = \theta(\sin(\ln \theta) + \cos(\ln \theta)) \). It is expressed in terms of \( \theta \). Our task is to find \( \frac{dy}{d\theta} \). This will require applying both the product rule and the chain rule of differentiation.

Step 2: Apply the Product Rule

Let's denote \( u(\theta) = \theta \) and \( v(\theta) = \sin(\ln \theta) + \cos(\ln \theta) \). The derivative \( \frac{dy}{d\theta} \) can be found using the product rule: \( \frac{d}{d\theta}(uv) = u'v + uv' \). First, we calculate \( u' = \frac{d}{d\theta}(\theta) = 1 \).

Step 3: Differentiate Inside the Bracket

We need \( v'(\theta) \). Start by differentiating each term inside the bracket. Use the chain rule for these derivatives. The derivative of \( \sin(\ln \theta) \) with respect to \( \theta \) is \( \cos(\ln \theta) \cdot \frac{d}{d\theta}(\ln \theta) = \frac{\cos(\ln \theta)}{\theta} \). Similarly, the derivative of \( \cos(\ln \theta) \) is \( -\sin(\ln \theta) \cdot \frac{d}{d\theta}(\ln \theta) = -\frac{\sin(\ln \theta)}{\theta} \).

Step 4: Combine the Derived Components for v'(θ)

Combine these results: \( v'(\theta) = \frac{\cos(\ln \theta)}{\theta} - \frac{\sin(\ln \theta)}{\theta} = \frac{1}{\theta}(\cos(\ln \theta) - \sin(\ln \theta)) \).

Step 5: Plug Components into Product Rule

Substitute \( u, u', v, \) and \( v' \) back into the product rule formula: \( \frac{dy}{d\theta} = 1 \cdot (\sin(\ln \theta) + \cos(\ln \theta)) + \theta \cdot \frac{1}{\theta}(\cos(\ln \theta) - \sin(\ln \theta)) \).

Step 6: Simplify the Expression

Simplify: \( \frac{dy}{d\theta} = \sin(\ln \theta) + \cos(\ln \theta) + \cos(\ln \theta) - \sin(\ln \theta) \). This simplifies to \( 2\cos(\ln \theta) \).

Step 7: Final Result

Thus, the derivative of \( y \) with respect to \( \theta \) is \( \frac{dy}{d\theta} = 2\cos(\ln \theta) \). Confirm the differentiation rules and arithmetic were applied correctly.

Key Concepts

Product Rule

The product rule is a key tool in differentiation, particularly when dealing with functions expressed as the product of two separate parts. If you have a function that is a product of two functions, say \( u(\theta) \) and \( v(\theta) \), the product rule tells us how to find the derivative of this product. The formula is \( \frac{d}{d\theta}(u v) = u'v + uv' \). This means we first differentiate \( u \) to get \( u' \) and then multiply it with the original \( v \). Next, we differentiate \( v \) to get \( v' \) and multiply it with the original \( u \). Finally, we add these two products together.
In our exercise, we apply this rule for \( y = \theta(\sin(\ln \theta) + \cos(\ln \theta)) \). Here, \( u(\theta) = \theta \) and \( v(\theta) = \sin(\ln \theta) + \cos(\ln \theta) \). We find that \( u' \) is simply 1 because the derivative of \( \theta \) with respect to \( \theta \) is 1. The challenge comes in differentiating \( v(\theta) \), which we'll tackle next using the chain rule.

Chain Rule

The chain rule is fundamental when differentiating composite functions, which are functions nested within other functions. It is used when you need to take the derivative of a function like \( \sin(\ln \theta) \).
To apply the chain rule, identify the "outer" function and the "inner" function. For \( \sin(\ln \theta) \), \( \sin \) is the outer function and \( \ln \theta \) is the inner function.

• First, differentiate the outer function with respect to the inner function, resulting in \( \cos(\ln \theta) \).
• Then, differentiate the inner function (\( \ln \theta \)) with respect to \( \theta \), resulting in \( 1/\theta \).

Multiply these derivatives together as per the chain rule to get the full derivative: \( \cos(\ln \theta) \cdot (1/\theta) = \frac{\cos(\ln \theta)}{\theta} \).
Similarly, applying the chain rule to \( \cos(\ln \theta) \), you first differentiate \( \cos \) to get \(-\sin \), then multiply by the inner derivative \( 1/\theta \): \(-\sin(\ln \theta) \cdot (1/\theta) = -\frac{\sin(\ln \theta)}{\theta} \).
This process allows us to solve for \( v'(\theta) \) effectively, which is critical for using the product rule fully.

Trigonometric Functions

Trigonometric functions often appear in calculus problems, adding a layer of complexity to differentiation tasks. In this exercise, functions such as \( \sin(x) \) and \( \cos(x) \) play a pivotal role. Understanding their derivatives is essential.
When it comes to these functions:

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

These derivatives are simple yet powerful tools in calculus.
Incorporating them into the differentiation process means recognizing how composite functions affect the basic form. When these functions have a more complex argument than just \( x \) (like \( \ln \theta \)), we use the chain rule to differentiate with respect to the underlying variable.
In our specific problem, \( \sin(\ln \theta) \) and \( \cos(\ln \theta) \) required careful application of differentiation rules to map back to the derivatives of simple sinusoidal functions, culminating in the final derivative expression. Knowing how trigonometric functions behave under differentiation is crucial for this type of calculus operation.