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The chain rule is a powerful technique in calculus used to differentiate composite functions. A composite function is a function that is made up of two or more simpler functions. In the context of the given exercise, the function is composed of the arcsine function and a square root function.To apply the chain rule, first identify the outer function and the inner function:

• The outer function is the arcsine function, \( F(u) = \arcsin(u) \).
• The inner function is the square root, \( u = \sqrt{\sin \theta} \).

The chain rule states that the derivative of a composite function \( F(g(x)) \) is given by \( F'(g(x)) \cdot g'(x) \). In other words, you need to multiply the derivative of the outer function, evaluated at the inner function, by the derivative of the inner function. This rule allows us to differentiate the function \( F(\theta) = \arcsin(\sqrt{\sin \theta}) \) efficiently.

The arcsine function, denoted as \( \arcsin(x) \), is the inverse of the sine function. It takes a value from the range \([-1, 1]\) and provides an angle in radians from \([-\frac{\pi}{2}, \frac{\pi}{2}]\).Differentiating the arcsine function with respect to its input, \( u \), requires knowing its derivative formula. The derivative of \( \arcsin(u) \) is given by:\[ \frac{d}{du}[\arcsin(u)] = \frac{1}{\sqrt{1-u^2}} \]This derivative formula only holds when \( |u| \leq 1 \). It helps us change angle values obtained from sine back to their original form before sine was applied. In the problem, we use this derivative when applying the chain rule. Understanding how to differentiate inverse trigonometric functions like \( \arcsin \) is crucial in solving calculus problems that involve complex functions.

Differentiation is the process of finding the derivative of a function, which represents how a function is changing at any point. In this exercise, we are interested in differentiating a function composed of the arcsine and square root functions.To differentiate the given function \( F(\theta) = \arcsin(\sqrt{\sin \theta}) \), we follow these steps:

• Differentiate the outer function: Using the derivative of arcsine, we get \( \frac{1}{\sqrt{1-u^2}} \), where \( u = \sqrt{\sin \theta} \).
• Differentiate the inner function: The inner function can be written as \( u = (\sin \theta)^{1/2} \). Its derivative, using the power rule and chain rule, is \( \frac{\cos \theta}{2\sqrt{\sin \theta}} \).

Combining these results through the chain rule gives the overall derivative. Differentiating composite functions often involves multiple rules and steps, making it an essential practice for mastering calculus.

Simplification in calculus involves rewriting expressions to make them clearer or easier to work with. In differentiation problems, simplification can help in both the derivation process and in presenting the final answer.After applying differentiation via the chain rule, the derivative of the function \( F(\theta) = \arcsin(\sqrt{\sin \theta}) \) is:\[ \frac{\cos \theta}{2\sqrt{\sin \theta} \cdot \sqrt{1-\sin \theta}} \]This result can be further simplified:

• The expression \( \sqrt{1 - (\sqrt{\sin \theta})^2} \) simplifies to \( \sqrt{1-\sin \theta} \).
• Finally, we can write the entire derivative expression as \( \frac{\cos \theta}{2\sqrt{\sin \theta(1-\sin \theta)}} \).

Simplification is crucial because it often unveils more about the behavior of the function and can make further calculus operations easier to perform. Practicing simplification ensures equations remain manageable and understandable.