Y=sqrt ((x-5)/(x-1)) найти y'(-1/3)

Y=√((x-5)/(x-1))
y' = ( 1 / 2√((x-5)/(x-1)) )* ((x-5)/(x-1))' =
(1 / 2) * √((x-1)/(x-5)) * ((x-5)/(x-1))' = A

((x-5)/(x-1))' =
((x-5)' * (x-1) - (x-5) * (x-1)' ) / (x-1)^2 =
(((x)'-(5)') * (x-1) - (x-5) * ((x)'-(1)') ) / (x-1)^2 =
((1-0) * (x-1) - (x-5) * (1-0) ) / (x-1)^2 =
(1 * (x-1) - (x-5) * 1) / (x-1)^2 =
(x-1 - x+5) / (x-1)^2 =
(-1 +5) / (x-1)^2 =
4 / (x-1)^2

f'(-1/3) = 4 / (-1/3 - 1)^2 =
4 / (-1/3 - 3/3)^2 =
4 / (-4/3)^2 =
4 / (16/9) =
(4*9)/16 =
9/4