# Let `G(x) = 2sin^(-1)((sqrt(x))/(2))` 1) Find G'(x). 2) The domain of G(x)is: 3) The domain of G'(x) is:

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`G(x)=2sin^(-1)(sqrtx/2)`

(1) To determine G'(x), use the formula `d/(du) sin^(-1)u= 1/sqrt(1-u^2)*u'` .

`G'(x)= 2*1/sqrt(1-(sqrtx/2)^2)*(sqrtx/2)'`

`G'(x)=2/sqrt(1-x/4)*1/2*1/(2sqrtx)`

`G'(x)=1/(2sqrtxsqrt((4-x)/4))`

`G'(x)=1/(sqrtxsqrt(4-x))`

**>> Hence, `G'(x) =1/(sqrtxsqrt(4-x)) ` .**

(2) To determine the domain of G(x), we need to take note the domain of the basic inverse sine function.

For `y=sin^(-1) x` , domain is `-1lt=xlt=1` .

So the domain of `G(x) = sin^(-1)(sqrtx/2)` should satisfy the condition `-1lt=sqrtx/2lt=1` .

To solve for x, set

`sqrtx/2 >=-1` and `sqrtx/2lt=1`

`sqrtxgt=-2` `sqrtxlt=2`

`xlt=4 ` `xlt=4`

Also for square roots, the radicand shouls be positive So,the values of x in `sqrtx` should be `x>=0` .

**>> Hence, the domain of G(x) is `0<=x<=4` .**

(3) To determine the domain of G'(x), we need to take note that in rational functions, we can not have a zero denominator.

Moreover, the denominator of G'(x) contains square roots. In square root, a negative radicand is not allowed.

Since we cannot have a zero denominator and a negative radicand, set each radicand in the denominator greater than zero.

`4-x>0` and `x > 0`

`-x > -4`

`x < 4`

>> **Hence, the domain of G'(x) is `0ltxlt4` .**