Problem 33 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ f(\theta)=\left(\frac{\sin \theta}{1+\cos \theta}\right)^{2} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{2 \sin \theta}{(1 + \cos \theta)^2} \).

Step 1: Understand the Function Composition

The function is given as \( f(\theta) = \left( \frac{\sin \theta}{1 + \cos \theta} \right)^2 \). We need to identify the outer and inner functions. Here, the outer function is \( u^2 \) and the inner function is \( u = \frac{\sin \theta}{1 + \cos \theta} \).

Step 2: Differentiate the Outer Function

Differentiate the outer function \( u^2 \) with respect to \( u \) to get \( \frac{d}{du} (u^2) = 2u \). This will be used in the chain rule application.

Step 3: Differentiate the Inner Function

Differentiate the inner function \( \frac{\sin \theta}{1 + \cos \theta} \) using the quotient rule. The quotient rule is given by \( \frac{d}{d\theta} \left( \frac{f(\theta)}{g(\theta)} \right) = \frac{g(\theta)f'(\theta) - f(\theta)g'(\theta)}{(g(\theta))^2} \).

Step 4: Apply the Quotient Rule to the Inner Function

Using the quotient rule, let \( f(\theta) = \sin \theta \) and \( g(\theta) = 1 + \cos \theta \). Then \( f'(\theta) = \cos \theta \) and \( g'(\theta) = -\sin \theta \). Applying the rule: \[ \frac{d}{d\theta} \left( \frac{\sin \theta}{1 + \cos \theta} \right) = \frac{(1 + \cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta)}{(1 + \cos \theta)^2} = \frac{\cos \theta + \cos^2 \theta + \sin^2 \theta}{(1 + \cos \theta)^2} \]

Step 5: Simplify the Inner Function's Derivative

Since \( \cos^2 \theta + \sin^2 \theta = 1 \), simplify to get \( \frac{\cos \theta + 1}{(1 + \cos \theta)^2} = \frac{1}{1 + \cos \theta} \).

Step 6: Apply the Chain Rule

Using the chain rule, combine the outer derivative and inner derivative: \[ \frac{df}{d\theta} = 2 \left( \frac{\sin \theta}{1 + \cos \theta} \right) \cdot \frac{1}{1 + \cos \theta} \]

Step 7: Simplify the Derivative Expression

Simplify the expression: \[ \frac{df}{d\theta} = \frac{2 \sin \theta}{(1 + \cos \theta)^2} \]

Step 8: Conclusion

The derivative of the function \( f(\theta) = \left( \frac{\sin \theta}{1 + \cos \theta} \right)^2 \) is \( \frac{2 \sin \theta}{(1 + \cos \theta)^2} \).

Key Concepts

Chain Rule

To understand the differentiation of complex functions, the chain rule is an essential concept. The chain rule allows us to differentiate a function composed of other functions. Think of it as peeling an onion; you remove layer by layer. In the expression \( f(\theta) = \left( \frac{\sin \theta}{1 + \cos \theta} \right)^2 \), we have an outer function, \( u^2 \), and an inner function, \( u = \frac{\sin \theta}{1 + \cos \theta} \).
Using the chain rule, we first compute the derivative of the outer function regarding the inner one: \( \frac{d}{du}(u^2) = 2u \).
This gives us something crucial to work with. Next, we differentiate the inner function with respect to \( \theta \). The chain rule combines these derivatives to form the overall derivative. Once you understand the distinction between the outer and inner functions, the chain rule simplifies what could otherwise be a challenging differentiation problem.

• Identify inner and outer functions.
• Differentiate the outer function.
• Differentiate the inner function.
• Multiply the results for the final derivative using the chain rule.

Quotient Rule

The quotient rule is your tool of choice for differentiating ratios of two functions. It becomes especially handy when dealing with functions of the form \( \frac{f(\theta)}{g(\theta)} \). Given our function \( \frac{\sin \theta}{1 + \cos \theta} \), the quotient rule tells us how to find its derivative.
The formula for the quotient rule is \( \frac{d}{d\theta} \left( \frac{f(\theta)}{g(\theta)} \right) = \frac{g(\theta)f'(\theta) - f(\theta)g'(\theta)}{(g(\theta))^2} \).
This formula helps us differentiate the inner function in the exercise. It's crucial to remember the following steps:

• Differentiate the numerator \( f(\theta) = \sin \theta \) to get \( f'(\theta) = \cos \theta \).
• Differentiate the denominator \( g(\theta) = 1 + \cos \theta \) to get \( g'(\theta) = -\sin \theta \).
• Plug these derivatives into the quotient rule formula.
• Simplify the result.

The math can get a bit intense, but remember: patience and careful calculation will guide you to the correct derivative formula. The quotient rule is an incredibly powerful method in calculus, even though it might seem cumbersome at first.

Trigonometric Functions

In calculus, trigonometric functions like \( \sin \theta \) and \( \cos \theta \) often appear in differentiation problems. These functions have unique derivatives:

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

Understanding these basic derivatives is crucial when simplifying expressions in calculations, just like it was used in this function differentiation.
When you're differentiating a trigonometric expression, always keep these rules at the forefront of your mind. It will help you simplify problems sooner rather than later.
Plus, if you stumble upon \( \sin^2 \theta + \cos^2 \theta \), don't forget it equals 1. It's a fundamental trigonometric identity that we used in the original exercise to simplify

\( \cos^2 \theta + \sin^2 \theta \)

to 1. This is not only a time-saver but also helps in reducing the complexity of the algebra involved while differentiating trigonometric functions.