Derivative of y = (x-1)^1/3 / (x+1)^1/3?

Use the quotient rule, since product rule would be too complicated

Let u = (x-1)^(1/3)
du/dx = 1/3(x-1)^(1/3-1)(1)
..........= 1/3(x-1)^(-2/3)

and let v= (x+1)^(1/3)
dv/dx = 1/3(x+1)(1/3-1)(1)
.........= 1/3(x+1)(-2/3)

By quotient rule, dy/dx = [v(du/dx) - u(dv/dx)]/v^2

v(du/dx) = [(x+1)^(1/3)][1/3(x-1)^(-2/3)]
..............= (∛x^2 -1)/(3x-3)

u(dv/dx) = [(x-1)^(1/3)][1/3(x+1)(-2/3)]
..............= (∛x^2 -1)/(3x + 3)

v^2 = [(x+1)^(1/3)]^2
......= ∛(x+1)^2


[v(du/dx) - u(dv/dx)]/v^2 = [(∛x^2 -1)/(3x-3) - (∛x^2 -1)/(3x + 3)]/∛(x+1)^2
........................................= [(3x+3)(∛x^2 -1) - (3x-3)(∛x^2 -1)/(3x-3)(3x+3)]/∛(x+1)^2
........................................= [(3x+3)(∛x^2 -1) - (3x-3)(∛x^2 -1)/(9x^2 -9)]/∛(x+1)^2

= .....

Edit: Looks like quotient rule is complicated too. Here: http://imgur.com/a/TJPcX

Please read :

i) Given: y = ∛{(x - 1)/(x + 1)}

ii) Taking log to base e on both sides,
ln|y| = (1/3)*[ln|x - 1| - ln|x + 1|]

iii) Differentiating both sides with respect to x,
(1/y)*(dy/dx) = (1/3)*[1/(x - 1) - 1/(x + 1)] = (1/3)*[(x + 1 - x + 1)/{(x - 1)(x + 1)}]

==> (1/y)*(dy/dx) = (1/3)*[2/(x² - 1)]

So, dy/dx = 2y/{3(x² - 1)}

==> dy/dx = [2*∛{(x - 1)/(x + 1)]/{3(x² - 1)}

y = (x - 1)^(1/3) / (x + 1)^(1/3)
y = ((x - 1) / (x + 1))^(1/3)
y = ((x + 1 - 2) / (x + 1))^(1/3)
y = ((x + 1) / (x + 1) - 2 / (x + 1))^(1/3)
y = (1 - 2 / (x + 1))^(1/3)
y^3 = 1 - 2 / (x + 1)
y^3 = 1 - 2 * (x + 1)^(-1)
3y^2 * dy = 0 - 2 * (-1) * (x + 1)^(-2) * dx
3y^2 * dy = 2 * (x + 1)^(-2) * dx
dy/dx = 2 / (3 * y^2 * (x + 1)^2)
y' = 2 / (3 * ((x - 1) / (x + 1))^(2/3) * (x + 1)^2)
y' = 2 / (3 * (x - 1)^(2/3) * (x + 1)^(2 - 2/3))
y' = 2 / (3 * (x - 1)^(2/3) * (x + 1)^(4/3))
y' = 2 * (x + 1)^(2/3) / (3 * (x - 1)^(2/3) * (x + 1)^2)
y' = 2 * (x + 1)^(2/3) * (x - 1)^(1/3) / (3 * (x - 1) * (x + 1)^2)
y' = 2 * ((x + 1)^2 * (x - 1))^(1/3) / (3 * (x - 1) * (x + 1)^2)
y' = 2 / (3 * ((x - 1) * (x + 1)^2)^(1/3))
y' = 2 / (3 * ((x^2 - 1) * (x + 1))^(1/3))
y' = 2 / (3 * (x^3 + x^2 - x - 1)^(1/3))