Problem 40 - University Calculus: Early Transcendentals

Question

Find the derivatives of the function. $$r=e^{\theta}\left(\frac{1}{\theta^{2}}+\theta^{-\pi / 2}\right)$$

Step-by-Step Solution

Answer

The derivative is \( r' = e^\theta \left(-2\theta^{-3} - \frac{\pi}{2}\theta^{-\pi/2 - 1}\right) + e^\theta \left(\frac{1}{\theta^2} + \theta^{-\pi/2}\right) \).

Step 1: Understand the function

The given function is \( r = e^\theta \left( \frac{1}{\theta^2} + \theta^{-\pi/2} \right) \). This function combines an exponential term, \( e^\theta \), with a sum of two power terms inside the parentheses.

Step 2: Apply the product rule

The function \( r \) involves a product of \( e^\theta \) and \( \left( \frac{1}{\theta^2} + \theta^{-\pi/2} \right) \). To differentiate, we need to apply the product rule. The product rule states \( (uv)' = u'v + uv' \), where \( u = e^\theta \) and \( v = \frac{1}{\theta^2} + \theta^{-\pi/2} \).

Step 3: Differentiate u

For \( u = e^\theta \), the derivative is straightforward. The derivative of \( e^\theta \) with respect to \( \theta \) is \( u' = e^\theta \).

Step 4: Differentiate v

For \( v = \frac{1}{\theta^2} + \theta^{-\pi/2} \), differentiate each term separately. The first term, \( \frac{1}{\theta^2} \), can be written as \( \theta^{-2} \). Its derivative is \( -2\theta^{-3} \). The derivative of the second term, \( \theta^{-\pi/2} \), is \( -\frac{\pi}{2}\theta^{-\pi/2 - 1} \).

Step 5: Combine using the product rule

Now apply the product rule: \( r' = e^\theta \left(-2\theta^{-3} - \frac{\pi}{2}\theta^{-\pi/2 - 1}\right) + e^\theta \left(\frac{1}{\theta^2} + \theta^{-\pi/2}\right) \). This can be simplified, if necessary, for clarity.

Key Concepts

Exponential Functions

Exponential functions are a fundamental concept in calculus and are characterized by the presence of a constant base raised to a variable exponent. In our case, the exponential function is represented by \( e^\theta \), where \( e \) is Euler's number (approximately 2.71828), a constant frequently used in mathematics due to its unique properties. The function \( e^x \) has a remarkable feature: its derivative is itself, \( \frac{d}{dx}e^x = e^x \). This property makes it very efficient to work with in calculus, especially in derivatives and integrals.
When solving problems involving exponential functions, it's crucial to recognize that regardless of the changes in the variable \( \theta \), the rate of growth is inherently determined by the function itself. This stability and predictiveness are why exponential functions are often used to model real-world phenomena like population growth and radioactive decay. Understanding these characteristics helps in analyzing how the function behaves and evolves.

Product Rule

The product rule is a powerful tool in calculus for differentiating functions that are products of two or more smaller functions. When you have two functions multiplied together, you cannot assume their derivative is simply the product of their individual derivatives. Instead, the product rule must be applied. The rule itself states that for two functions \( u \) and \( v \), the derivative of their product is given by:

• \((uv)' = u'v + uv'\)

This means you first take the derivative of \( u \) and multiply it by \( v \), then take the derivative of \( v \) and multiply it by \( u \), and finally sum these results.In our exercise, \( u = e^\theta \) and \( v = \frac{1}{\theta^2} + \theta^{-\pi/2} \). Applying the product rule helps you find the derivative, which is essential for understanding how the overall function behaves. This concept is especially useful when working with complex equations, where direct differentiation isn't possible.

Power Rule

The power rule is a fundamental technique to differentiate functions with powers of the variable. It simplifies finding the derivative of polynomial forms, and is applicable to differentiating terms like \( \theta^n \). For a function \( f(x) = x^n \), where \( n \) is any real number, the power rule tells us:

• \( \frac{d}{dx}x^n = nx^{n-1} \)

This rule is straightforward: multiply the original power by the coefficient and decrement the power by one.In the provided exercise, we use this rule for both terms \( \theta^{-2} \) and \( \theta^{-\pi/2} \). When differentiating \( \theta^{-2} \), apply the power rule to obtain \( -2\theta^{-3} \). Similarly, for \( \theta^{-\pi/2} \), the derivative is \( -\frac{\pi}{2}\theta^{-\pi/2 - 1} \).The power rule is crucial in calculus because it allows you to break down and analyze functions term by term, making complex derivatives more manageable.