91Ó°ÊÓ

Implicit differentiation is a powerful tool, especially useful when dealing with equations where the dependent variable is not isolated on one side. In the context of finding the derivative of \( \sin^{-1}(x) \), we start with the relationship \( y = \sin^{-1}(x) \), implying that \( x = \sin(y) \). This is an implicit equation since \( y \) is contained within the function \( \sin(y) \) on the right-hand side.

To differentiate implicitly, we treat \( y \) as a function of \( x \) and differentiate both sides with respect to \( x \). Doing so gives us \( \frac{d}{dx} \sin(y) = \cos(y) \cdot \frac{dy}{dx} \). Applying this to the equation \( x = \sin(y) \), we differentiate the left side, resulting in 1 because the derivative of \( x \) with respect to itself is 1. The right side involves the chain rule, yielding \( \cos(y) \cdot \frac{dy}{dx} \).

Finally, we solve for \( \frac{dy}{dx} \), resulting in \( \frac{dy}{dx} = \frac{1}{\cos(y)} \). The challenge remains to express this in terms of \( x \), necessitating a smart substitution using trigonometric identities.

The Pythagorean identity is a crucial trigonometric identity utilized to express trigonometric functions in terms of others. It is stated as \( \sin^2(y) + \cos^2(y) = 1 \). This identity helps us connect \( \sin(y) \) and \( \cos(y) \) when working with inverse trigonometric functions.

In Step 4 of finding the derivative of \( \sin^{-1}(x) \), we have established \( x = \sin(y) \). Thus, \( \sin^2(y) = x^2 \). Substituting this into the Pythagorean identity, we solve for \( \cos^2(y) \):

• \( \sin^2(y) = x^2 \)
• \( \cos^2(y) = 1 - x^2 \)
• \( \cos(y) = \sqrt{1 - x^2} \)

Given \( \cos^2(y) = 1 - x^2 \), taking the square root helps us express the result \( \cos(y) = \sqrt{1-x^2} \), which is necessary to substitute back in the implicit differentiation formula.

Inverse trigonometric functions are designed to yield an angle whose trigonometric function would produce a given value. For \( \sin^{-1}(x) \), it means finding the angle \( y \) such that \( \sin(y) = x \). The domain and range for inverse functions are critical too, ensuring that they are well-defined and meaningful.

When dealing with the derivative of an inverse trigonometric function, like \( \sin^{-1}(x) \), we think about the function's behavior within its allowable range, typically \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \) for arcsine, and within the domain \(-1 \leq x \leq 1\). The process of implicit differentiation helps bridge the gap between the derivative of the function and its inverse.

Applying these concepts, the derivative of \( \sin^{-1}(x) \) is found as \( \frac{1}{\sqrt{1-x^2}} \). This concise result is useful for integrating and understanding arcsine's rate of change with respect to \( x \). It's important to remember this derivative result can only be applied directly when \( x \) is within \(-1, 1\).