To find the second derivative of y with respect to x, we need to differentiate y' with respect to x.

Given: y = 2 sin^(-1)(cosx)

Let's first find the first derivative of y (y') and then differentiate it again to find the second derivative (d^2y/dx^2).

First Derivative (y'):
To find y', we need to use the chain rule. Let's differentiate each part of the expression step by step:

1. Differentiate the outer function: sin^(-1)(cosx)
The derivative of sin^(-1)(u) is 1/√(1 - u^2), where u is the function inside the inverse sine.

So, the derivative of sin^(-1)(cosx) with respect to cosx is 1/√(1 - cos^2x).

2. Differentiate the inner function: cosx
The derivative of cosx with respect to x is -sinx.

3. Apply the chain rule:
Multiply the derivative of the outer function with respect to the inner function (1/√(1 - cos^2x)) and the derivative of the inner function with respect to x (-sinx).

Therefore, y' = (1/√(1 - cos^2x)) * (-sinx)
Simplifying, y' = -sinx/√(1 - cos^2x)

Second Derivative (d^2y/dx^2):
To find the second derivative, we need to differentiate y' with respect to x.

1. Differentiate y' = -sinx/√(1 - cos^2x)
To differentiate y', we can use the quotient rule.

The quotient rule states that if we have a function of the form f(x)/g(x), then the derivative is given by [(g(x)f'(x) - f(x)g'(x))/g(x)^2].

Let's differentiate y' using the quotient rule:

f(x) = -sinx
f'(x) = -cosx

g(x) = √(1 - cos^2x)
g'(x) = (1/2) * (1 - cos^2x)^(-1/2) * (-2cosx) = -cosx/√(1 - cos^2x)

Applying the quotient rule:
y'' = [(g(x)f'(x) - f(x)g'(x))/g(x)^2]
= [(-cosx/√(1 - cos^2x)) * (-cosx) - (-sinx) * (-cosx/√(1 - cos^2x))]/(√(1 - cos^2x))^2
= [(cos^2x/√(1 - cos^2x)) + (sinx * cosx/√(1 - cos^2x))]/(1 - cos^2x)
= [cos^2x + sinx * cosx]/[(1 - cos^2x)√(1 - cos^2x)]

Simplifying the expression further is a bit challenging, but we can see that the expression for y'' is quite complicated. However, if we substitute cos^2x with 1 - sin^2x (using the identity cos^