Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 41

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)