To take the derivative of the given function:` g(x) =3arccos(x/2)` ,

we can apply the basic property: `d/(dx) [c*f(x)] = c * d/(dx) [f(x)]` .

then `g'(x) = 3 d/(dx) (arccos(x/2))`

To solve for the `d/(dx) (arccos(x/2))` , we consider the derivative formula of an inverse trigonometric function.

For the derivative of inverse "cosine" function, we follow:

`d/(dx) (arccos(u)) = -((du)/(dx))/sqrt(1-u^2)`

To apply the formula with the given function, we let `u= x/2` then` (du)/(dx) = 1/2` .

`d/(dx) (arccos(x/2))= - (1/2)/sqrt(1-(x/2)^2)`

Evaluate the exponent:

`d/(dx) (arccos(x/2))= - (1/2)/sqrt(1-x^2/4)`

Express the expression inside radical as one fraction:

`d/(dx) (arccos(x/2))= - (1/2)/sqrt((4-x^2)/4)`

Apply the property of radicals: `sqrt(a/b)= sqrt(a)/sqrt(b)` at the bottom:

`d/(dx) (arccos(x/2))= - (1/2)/((sqrt(4-x^2)/sqrt(4)))`

To simplify, flip the bottom to proceed to multiplication:

`d/(dx) (arccos(x/2))= - (1/2)* sqrt(4)/sqrt((4-x^2))`

`d/(dx) (arccos(x/2))= - (1/2)* 2/sqrt((4-x^2))`

Multiply across:

`d/(dx) (arccos(x/2))= - 2/(2sqrt(4-x^2))`

Cancel out the common factor 2 from top and bottom:

`d/(dx) (arccos(x/2))= - 1/sqrt(4-x^2)`

With `d/(dx) (arccos(x/2))= - 1/sqrt(4-x^2)` , then

`g'(x) = 3 *d/(dx) (arccos(x/2))`

becomes

`g'(x)= 3*- 1/sqrt(4-x^2)`

`g'(x)=-3/sqrt(4-x^2)`