trigonometric function

y = 2sin^-1(x-1)

 

Divide both sides by 2 and take sin of both sides of the result

sin(y/2) = x - 1      ...(1)

 

d(sin(y/2))/dx = d(x-1)/dx

d(sin(y/2))/dy*dy/dx = d(x-1)/dx    using the chain rule on the LHS

 

(1/2)*cos(y/2)*dy/dx = 1     performing the differentiation on each side

 

Rearrange:

dy/dx = 2/cos(y/2)

dy/dx = 2/√(1-sin²(y/2))

 

Make use of (1)
dy/dx = 2/√(1 - (x-1)²)