Let F(x) = arcsin((x-2)/(2)) 1) Find F'(x).   2) Find The domain of F(x)   3) Find the domain of F'(x)  

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1) You need to differentiate the given function with respect to x, using the chain rule, such that:

`F'(x) = (1/(sqrt(1 - (x-2)^2/4)))*((x-2)/2)'`

`F'(x) = 2/(2sqrt(4 - (x-2)^2)) => F'(x) = 1/((2-x+2)(2+x-2))`

`F'(x) = 1/(x(4-x))`

Hence, differentiating the given function yields `F'(x) = 1/(x(4-x)).`

2) You should remember that the domain of the arcsine function is `[-1,1], ` hence `(x-2)/2 in [-1,1]`  such that:

`-1 =< (x-2)/2 =< 1 => -2 =< x - 2 =< 2`

`0 =< x =< 4`

Hence, evaluating the domain of the function F(x) yields `x in [0,4].`

3) Notice that the function F'(x) is a fraction, hence its denominator needs to be different from zero such that:

`x(4 - x)!=0 => {(x!=0),(4-x!=0):} => {(x!=0),(x!=4):}`

Hence, evaluating the domain of the given function yields`x in R - {0 ; 4}.`

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