91Ó°ÊÓ

When dealing with derivatives, the quotient rule is a powerful tool for functions that are presented as a division of two other functions. In our example, the function is \[ T = \frac{R}{\sin^{-1}(3R)} \]where the numerator is \( R \), and the denominator is \( \sin^{-1}(3R) \). The quotient rule helps us find the derivative of this division. It's like having a recipe that tells us how to mix up the derivatives correctly to get the result.
According to the quotient rule, for a function \( \frac{u}{v} \), the derivative is:

• \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \)

Here,

• \( u = R \) and so \( u' = 1 \) (since the derivative of \( R \) with respect to \( R \) is 1)
• \( v = \sin^{-1}(3R) \) (an inverse trigonometric function we will tackle next)

This tool, the quotient rule, is essential when directly differentiating a fraction-like expression where both the top and bottom could change with respect to the same variable.

The chain rule is like a stepping stone for differentiating composed functions—functions inside of other functions. In our case, the denominator of our given exercise involves an inverse trigonometric function of \( 3R \):
\( v = \sin^{-1}(3R) \). To find its derivative, we use the chain rule because \( \sin^{-1}(3R) \) is a function inside of another function. It's like peeling an onion; you tackle one layer at a time.
For the outside function \( \sin^{-1}(x) \), the derivative is \( \frac{1}{\sqrt{1-x^2}} \). The chain rule states:

• First, take the derivative of the outer function: \( \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}} \)
• Then multiply by the derivative of the inner function \( 3R \), which is 3: \( \frac{d}{dR}(3R) = 3 \)

Thus, the derivative of \( \sin^{-1}(3R) \) becomes \( \frac{3}{\sqrt{1-(3R)^2}} = \frac{3}{\sqrt{1-9R^2}} \).
The chain rule is all about consecutive actions, handling complex nestings of functions via their derivatives.

Inverse trigonometric functions help in navigating from a ratio or value back to an angle. In calculus, these functions such as \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \) hold importance because of their derivatives. They enable us to retrieve angle measures from trigonometric ratios.
For our exercise, the function \( \sin^{-1}(3R) \) is a classic example.

• The derivative \( \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}} \) becomes \( \frac{d}{dR}(\sin^{-1}(3R)) = \frac{3}{\sqrt{1-9R^2}} \) when we apply the chain rule.

These derivatives are pivotal; they correspond directly to the rate at which the angle changes compared to how fast \( R \) changes.
By understanding the derivatives of inverse trigonometric functions, you're equipped to analyze a broader class of problems, expanding your calculus toolkit.