Problem 25 - Thomas Calculus

Question

In Exercises \(5-36,\) 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 \( \frac{dy}{d\theta} = 2 \cos(\ln \theta) \).

Step 1: Identify the Differentiation Variable

The function given is in terms of \( \theta \). Therefore, we need to find \( \frac{dy}{d\theta} \).

Step 2: Apply the Product Rule

The function is of the form \( y = u \cdot v \) where \( u = \theta \) and \( v = \sin(\ln \theta) + \cos(\ln \theta) \). Apply the product rule: \( \frac{d}{d\theta}(u \cdot v) = u'v + uv' \).

Step 3: Differentiate \( u = \theta \)

Calculate \( u' = \frac{d}{d\theta}(\theta) = 1 \).

Step 4: Differentiate \( v = \sin(\ln \theta) + \cos(\ln \theta) \)

First, find \( \frac{d}{d\theta}\sin(\ln \theta) \) using the chain rule: \( \cos(\ln \theta) \cdot \frac{d}{d\theta}(\ln \theta) = \frac{1}{\theta}\cos(\ln \theta) \). Then, find \( \frac{d}{d\theta}\cos(\ln \theta) \) using the chain rule: \( -\sin(\ln \theta) \cdot \frac{1}{\theta} = -\frac{1}{\theta}\sin(\ln \theta) \). Combine these results: \( v' = \frac{1}{\theta}\cos(\ln \theta) - \frac{1}{\theta}\sin(\ln \theta) \).

Step 5: Substitute and Simplify Expression

Substitute \( u' = 1 \), \( u = \theta \), \( v \), and \( v' \) into the product rule formula: \( \frac{dy}{d\theta} = 1 \cdot (\sin(\ln \theta) + \cos(\ln \theta)) + \theta (\frac{1}{\theta}\cos(\ln \theta) - \frac{1}{\theta}\sin(\ln \theta)) \). Simplify this to get \( \sin(\ln \theta) + \cos(\ln \theta) + \cos(\ln \theta) - \sin(\ln \theta) = 2\cos(\ln \theta) \).

Step 6: Write Final Derivative

The final expression after simplification is \( \frac{dy}{d\theta} = 2\cos(\ln \theta) \).

Key Concepts

Product Rule

The product rule is a vital tool in differentiation. It is used when you have a function that is the product of two or more functions. To differentiate a product of functions, you need to take into account how each function changes individually.

Consider two functions, say \( u(x) \) and \( v(x) \). The product rule states that the derivative of their product \( y = u(x) \cdot v(x) \) is:

• \( \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \)

The rule essentially tells you to take the derivative of the first function and multiply it by the second function, then add it to the product of the first function and the derivative of the second function.

In our example, we applied the product rule to \( y = \theta(\sin(\ln \theta) + \cos(\ln \theta)) \). Here, \( u = \theta \) and \( v = \sin(\ln \theta) + \cos(\ln \theta) \). By following the product rule, we first found \( u' \) as \( 1 \), and later calculated \( v' \) using additional differentiation rules.

Chain Rule

The chain rule is used to differentiate composite functions, where one function is inside of another. This rule is essential when you're facing functions nested within other functions.

Suppose you have a composite function like \( h(x) = f(g(x)) \). According to the chain rule, the derivative of \( h(x) \) with respect to \( x \) is:

• \( h'(x) = f'(g(x)) \cdot g'(x) \)

By applying the chain rule, you're essentially peeling back the layers of functions, much like an onion, differentiating from the outside in.

In our example, we used the chain rule to find the derivatives of \( \sin(\ln \theta) \) and \( \cos(\ln \theta) \). For \( \sin(\ln \theta) \), the outer function is sine and the inner is natural logarithm. We found the derivative using the chain rule as \( \cos(\ln \theta) \cdot \frac{1}{\theta} \). Similarly, for the cosine term, we differentiated with a minus sign, as cosine's derivative is negative sine.

Trigonometric Differentiation

Trigonometric differentiation applies to functions involving sine, cosine, and other trigonometric functions. These functions often require different rules compared to polynomial or exponential functions.

Key derivatives to remember include:

• The derivative of sine is cosine: \( \frac{d}{dx}(\sin(x)) = \cos(x) \).
• The derivative of cosine is negative sine: \( \frac{d}{dx}(\cos(x)) = -\sin(x) \).

These basic rules help tackle more complex expressions involving trigonometric functions.

In our specific case, the derivatives of \( \sin(\ln \theta) \) and \( \cos(\ln \theta) \) involved applying these standard trigonometric differentiation rules, alongside the chain rule, due to the presence of the natural logarithm function inside the trigonometric functions.

Logarithmic Function Differentiation

Differentiation of logarithmic functions has its special rules. The natural logarithm, denoted \( \ln(x) \), follows a straightforward rule, making it easier to differentiate:

For a function \( y = \ln(x) \), its derivative is given by:

• \( \frac{dy}{dx} = \frac{1}{x} \)

This fundamental rule helps in tackling more complex expressions involving logarithms within other functions.

In our problem, we used this property of logarithmic differentiation in combination with the chain rule to find derivatives of functions like \( \sin(\ln \theta) \) and \( \cos(\ln \theta) \). The logarithmic part, \( \ln \theta \), was specifically tackled by multiplying the derivative of the outer function by \( \frac{1}{\theta} \), which is the derivative of \( \ln(\theta) \). This approach was key to solving the derivative of the original function efficiently.