Problem 22 - Essential Calculus Early Transcendentals

Question

Find the derivative of the function. Simplify where possible. $$F(\theta)=\arcsin \sqrt{\sin \theta}$$

Step-by-Step Solution

Answer

\( \frac{\cos(\theta)}{2\sqrt{\sin(\theta)} \sqrt{1-\sin(\theta)}} \) is the derivative.

Step 1: Apply the Chain Rule

We have a composite function \( F(\theta) = \arcsin(u) \) where \( u = \sqrt{\sin(\theta)} \). The derivative of \( \arcsin(u) \) is \( \frac{1}{\sqrt{1-u^2}} \). We will use the chain rule here to differentiate \( F(\theta) \).

Step 2: Differentiate the Inside Function

Differentiate \( u = \sqrt{\sin(\theta)} \) with respect to \( \theta \). We do this by applying the chain rule: \( u = (\sin(\theta))^{1/2} \), so \( \frac{du}{d\theta} = \frac{1}{2}(\sin(\theta))^{-1/2} \cdot \cos(\theta) \). This simplifies to \( \frac{du}{d\theta} = \frac{\cos(\theta)}{2\sqrt{\sin(\theta)}} \).

Step 3: Combine Using the Chain Rule

Combine the derivatives using the chain rule: \( \frac{dF}{d\theta} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{d\theta} \). Substitute \( u = \sqrt{\sin(\theta)} \) and its derivative from the previous step.

Step 4: Plug in Values and Simplify

With \( u = \sqrt{\sin(\theta)} \), we have \( 1-u^2 = 1 - \sin(\theta) \). Thus, \( \frac{1}{\sqrt{1-u^2}} = \frac{1}{\sqrt{1-\sin(\theta)}} \). The final expression is \( \frac{dF}{d\theta} = \frac{\cos(\theta)}{2\sqrt{\sin(\theta)} \sqrt{1-\sin(\theta)}} \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. It helps in finding the derivative of functions that are composed of two or more functions. In layman's terms, the chain rule says that if you have a function inside another function, you need to differentiate both and then multiply their derivatives.

For instance, in the original problem, you have a composite function, namely, the arcsin and square root functions combined. You differentiate both parts separately and combine their results using the chain rule. Essentially, break it down by

• Identifying the outer and inner functions
• Computing the derivative of the outer function with respect to the inner
• Then, computing the derivative of the inner function with respect to the variable
• Multiplying these results together

This is the step that simplifies the process of finding derivatives of complex functions.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function changes. This is a crucial part of calculus, as it helps understand how functions behave and change over time or space.

In the context of the exercise, differentiation tells us how the composite function involving the arc sine and square root changes concerning the angle \( \theta \). By applying the chain rule, you obtain the derivative.

The step-by-step differentiation might look a bit complex with formulas, but the simplified concept remains to identify, differentiate, and combine results. Each component of the function you differentiate relates directly to understanding how the composite function changes.

Composite Function

A composite function arises when you apply one function to the result of another function. It is essentially a function within another function. Recognizing composite functions is essential when you want to apply the chain rule for differentiation.

In the given problem, \( F(\theta) = \arcsin(\sqrt{\sin(\theta)}) \) is a typical example of a composite function, combining the arcsine and square root functions. The inner function is \( \sqrt{\sin(\theta)} \), and the outer function is \( \arcsin(u) \), where \( u \) represents the inner function.

To differentiate it, you need to:

• Identify each separate function
• Apply differentiation rules to each part
• Use the chain rule to combine them

Understanding composite functions helps map out this path, crucial for tackling more complex calculus problems.

Trigonometric Derivatives

Trigonometric derivatives are specific derivatives of trigonometric functions. They help in understanding how trigonometric functions like sine, cosine, tangent, and their inverses change. When these functions are part of a composite, knowing each derivative rules aids in finding the overall change rate.

In the problem, the function includes \( \sin(\theta) \), a trigonometric function. Its derivative, \( \cos(\theta) \), feeds into the differentiation of the inside expression \( \sqrt{\sin(\theta)} \).

When taking derivatives involving inverse trigonometric functions like arcsine, it's useful to remember:

• \( \frac{d}{dx}(\arcsin(x)) = \frac{1}{\sqrt{1-x^2}} \)

Blending these derivative rules with trigonometric expressions, like those in the exercise, gives a better grasp of the chain rule in action within these contexts.