Problem 46 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$r=\sec \sqrt{\theta} \tan \left(\frac{1}{\theta}\right)$$

Step-by-Step Solution

Answer

\( r' = \sec \sqrt{\theta} \tan \sqrt{\theta} \cdot \frac{1}{2\sqrt{\theta}} \tan \left(\frac{1}{\theta}\right) - \sec \sqrt{\theta} \sec^2 \left(\frac{1}{\theta}\right) \cdot \frac{1}{\theta^2} \).

Step 1: Understand the Expression

The given function is \( r = \sec \sqrt{\theta} \tan \left(\frac{1}{\theta}\right) \). This expression is a product of two separate functions: \( \sec \sqrt{\theta} \) and \( \tan \left(\frac{1}{\theta}\right) \). We will find the derivative using the product rule.

Step 2: Apply the Product Rule

The product rule states that the derivative of two functions \( u \) and \( v \) is \( u'v + uv' \). Here, let \( u = \sec \sqrt{\theta} \) and \( v = \tan \left(\frac{1}{\theta}\right) \). We need to find \( u' \) and \( v' \).

Step 3: Find the Derivative of \( u = \sec \sqrt{\theta} \)

To find \( u' \), use the chain rule. \( u = \sec \sqrt{\theta} \) implies \( u = \frac{1}{\cos \sqrt{\theta}} \), so we have:\[ u' = \sec \sqrt{\theta} \tan \sqrt{\theta} \cdot \frac{1}{2\sqrt{\theta}} \].This is derived from the derivative of \( \sec \) and the chain rule for \( \sqrt{\theta} \) as \( \frac{d}{d\theta}(\sqrt{\theta}) = \frac{1}{2\sqrt{\theta}} \).

Step 4: Find the Derivative of \( v = \tan \left(\frac{1}{\theta}\right) \)

Use the chain rule again. \( v = \tan \left(\frac{1}{\theta}\right) \) implies:\[ v' = \sec^2 \left(\frac{1}{\theta}\right) \cdot \left(-\frac{1}{\theta^2}\right) \].This is derived from the derivative of \( \tan(x) = \sec^2(x) \) and the derivative of \( \frac{1}{\theta} = -\frac{1}{\theta^2} \).

Step 5: Substitute into the Product Rule

Substitute \( u' \) and \( v' \) into the product rule:\[ r' = \left(\sec \sqrt{\theta} \tan \sqrt{\theta} \cdot \frac{1}{2\sqrt{\theta}}\right) \tan \left(\frac{1}{\theta}\right) + \sec \sqrt{\theta} \cdot \left(\sec^2 \left(\frac{1}{\theta}\right) \left(-\frac{1}{\theta^2}\right)\right). \]

Step 6: Simplify the Expression

Now, simplify the expression for \( r' \):\[ r' = \sec \sqrt{\theta} \tan \sqrt{\theta} \cdot \frac{1}{2\sqrt{\theta}} \tan \left(\frac{1}{\theta}\right) - \sec \sqrt{\theta} \sec^2 \left(\frac{1}{\theta}\right) \cdot \frac{1}{\theta^2}. \]

Key Concepts

Product Rule

The product rule is an essential tool in calculus used for finding the derivative of a product of two functions. When you have two functions, say \( u(x) \) and \( v(x) \), and you need to differentiate their product \( u(x) \cdot v(x) \), the product rule provides a straightforward method to do that.

The product rule states:

• The derivative of \( u \cdot v \) is given by \( u'v + uv' \).

This means you take the derivative of the first function, multiply it by the second function, and then add the product of the first function and the derivative of the second function.

In our exercise, where \( r = \sec \sqrt{\theta} \tan \left(\frac{1}{\theta}\right) \), we set \( u = \sec \sqrt{\theta} \) and \( v = \tan \left(\frac{1}{\theta}\right) \). Here, applying the product rule allows us to structure the differentiation process methodically. This rule is particularly handy when direct differentiation of the entire product is complex.

Chain Rule

The chain rule is another critical concept in calculus used for differentiating composite functions. A composite function is one made up by combining two or more functions. With the chain rule, you can find the derivative of such functions efficiently.

The chain rule formula is:

• If you have a function \( f(g(x)) \), its derivative is \( f'(g(x)) \cdot g'(x) \).

The idea is to first differentiate the outer function while keeping the inner function intact, then multiply by the derivative of the inner function.

In the original exercise, this rule is used twice:

• When differentiating \( \sec \sqrt{\theta} \), we handle \( \sqrt{\theta} \) as the inner function and \( \sec \) as the outer.
• When finding the derivative of \( \tan \left(\frac{1}{\theta}\right) \), \( \frac{1}{\theta} \) becomes the inner function.

These compositions make the chain rule indispensable for solving such problems effectively. It aids in breaking down complex expressions into manageable parts.

Trigonometric Functions

Trigonometric functions such as \( \sec \), \( \tan \), and others play a crucial role in calculus, especially when taking derivatives. These functions are periodic and exhibit specific behaviors, which significantly influence how their derivatives behave.

Within the exercise, two primary trigonometric functions are involved:

• \( \sec(x) = \frac{1}{\cos(x)} \)
• \( \tan(x) = \frac{\sin(x)}{\cos(x)} \)

To find their derivatives:

• The derivative of \( \sec(x) \) is \( \sec(x) \tan(x) \). This arises from applying the quotient rule since \( \sec \) is inherently a quotient of \( 1 \) over \( \cos(x) \).
• The derivative of \( \tan(x) \) is \( \sec^2(x) \). Recognizing that \( \tan(x) \) represents the ratio \( \sin(x)/\cos(x) \) helps in understanding why this is surprisingly uncomplicated compared to its base terms.

Understanding these derivatives is essential for effectively applying them in calculus problems. They can simplify the work by transforming complex expressions step by step.

Derivative

The concept of a derivative is fundamental in calculus and represents the rate of change of a function. In simpler terms, it describes how a function changes as its input changes. Derivatives help in understanding the slope of a curve at any given point.

For a function \( f(x) \), the derivative is denoted as \( f'(x) \) or \( \frac{df}{dx} \). This notation helps signify that the derivative is measuring how \( f \) changes with a small change in \( x \).

In the context of our given expression, finding \( r' \), the derivative of \( r \), gives insight into how the trigonometric function behaves with changing \( \theta \).

This is not just a mechanical process; it offers a deeper understanding of the function's behavior:

• If \( r' \) is positive, the function \( r \) increases as \( \theta \) increases.
• If \( r' \) is negative, the function \( r \) decreases as \( \theta \) increases.

Recognizing the derivation of the expression's growth or decay is crucial in both theoretical and practical applications, such as physics or engineering.