91Ó°ÊÓ

The chain rule is a fundamental concept in calculus used to differentiate composite functions. When you have a function within another function, the chain rule helps to find the derivative of the entire expression efficiently. For example, if you have a function comprised of two parts, like this: \( y = f(g(x)) \), the chain rule states that the derivative of \( y \) with respect to \( x \) is the derivative of \( f \) with respect to \( g(x) \) multiplied by the derivative of \( g(x) \) with respect to \( x \). This can be written as:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

In the exercise, the chain rule is applied to differentiate \( y = \sqrt{\sin^{-1}(x-1)} \), where you identify the outer function as the square root and the inner function as the inverse sine.

Composition of functions involves creating a new function by applying one function to the result of another. This is often written as \( (f \circ g)(x) = f(g(x)) \). In the given exercise, the function \( y = \sqrt{\sin^{-1}(x-1)} \) is a composition of two functions: the square root function and the inverse sine function.

• Outer function: \( f(u) = \sqrt{u} \)
• Inner function: \( g(x) = \sin^{-1}(x-1) \)

By recognizing this structure, it becomes easier to apply the chain rule, where you first differentiate the outer function with respect to its input, and then differentiate the inner function with respect to \( x \). This approach systematically breaks down complex expressions into manageable pieces.

The square root function, \( f(x) = \sqrt{x} \), is one of the basic functions we often encounter and need to differentiate. The derivative for the function \( f(x) = \sqrt{x} \) is given by the formula:

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

This formula is used to find the rate of change of a square root function with respect to \( x \). In our problem, however, the function we need to differentiate, \( \sqrt{\sin^{-1}(x-1)} \), requires considering the square root as an outer function applied to the inverse sine function. First, find the derivative of the outer function as if the inner function was just \( u \) and then multiply it by the derivative of that inner function using the chain rule.

The inverse sine function, denoted by \( \sin^{-1}(x) \), also called arcsine, maps a real number between \(-1\) and \(1\) to an angle whose sine is that number. To find its derivative, we use the known derivative formula:

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

In the situation described in the exercise, the function is \( \sin^{-1}(x-1) \). All you have to do is apply this derivative rule by substituting \( x-1 \) instead of \( x \), resulting in:

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

This expression becomes part of the composite function's derivative calculation when paired with the derivative of the outer square root function.