Show that d/dx {[(2+x )/(2-x)^(1/2)]} = 2/[(2-x)^(3/2) * (2+x)^(1/2)] .
There's a slight mistake in your writing down the expression (2+x )/(2-x)^(1/2).
It should have been, [(2+x )/(2-x)]^(1/2). i.e the power (1/2) applies to both terms.
Let u = (2+x)^(1/2) and u' = (1/2)(1)(2+x)^(-1/2) = (1/2)(2+x)^(-1/2)
Let v = (2-x)^(1/2) and v' = (1/2)(-1)(2-x)^(-1/2) = (-1/2)(2 - x)^(-1/2)
f(x) = u/v
f' = (v.u' - u.v')/v^2
f' = {(2-x)^(1/2).(1/2).(2+x)^(-1/2) - (2+x)^(1/2).(-1/2).(2 - x)^(-1/2)}/(2-x)
f' = (1/2)(1/(2-x)){(2-x)^(1/2).(2+x)^(-1/2) + (2+x)^(1/2).(2-x)^(-1/2)}
f' = 1/(2(2-x)^(3/2)){(2-x)/(2+x)^(1/2) + (2+x)^(1/2)}
f' = 1/(2(2-x)^(3/2)){(2-x + 2-x)/(2+x)^(1/2)}
f' = 1/(2(2-x)^(3/2)){(4)/(2+x)^(1/2)}
df/dx = 2/[(2-x)^(3/2).(2+x)^(1/2)]