Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 39

The derivative of function f with respect to x is denoted as f'(x) .
To take the derivative of the given function: f(x) =2arcsin(x-1) ,
we can apply the basic property: d/(dx) [c*f(x)] = c * d/(dx) [f(x)] .
then f'(x) = 2 d/(dx) (arcsin (x-1))
To solve for the d/(dx) (arcsin(x-1)) , we consider the derivative formula of an inverse trigonometric function.
For the derivative of inverse "sine" function, we follow:
d/(dx) (arcsin (u)) = ((du)/(dx))/sqrt(1-u^2)

To apply the formula with the given function, we let u= x-1 then (du)/(dx) = 1 .
d/(dx) (arcsin(x-1))= 1/sqrt(1-(x-1)^2)
Then f'(x)=2 d/(dx) (arcsin (x-1)) becomes:
f'(x) =2 * 1/sqrt(1-(x-1)^2)
f'(x) =2 /sqrt(1-(x-1)^2)

To further simplify, we can evaluate the exponent inside the radical:
f'(x) =2/sqrt(1-(x^2-2x+1))
Note: (x-1)^2= (x-1)(x-1)
Applying FOIL or distributive property:
(x-1)(x-1)= x*x + x*(-1) + (-1)*x + (-1)(-1)
=x^2 –x –x+1
=x^2 -2x+1

Simplify the expression inside radical:
f'(x) =2/sqrt(1-x^2+2x -1)
f'(x)=2/sqrt(-x^2+2x) or f'(x)=2/sqrt(2x-x^2)