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)`
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