(t^2 - 1)^6 /t. Find f'(t).?

Question

(t(12((t^2)-1)^5))-(((t^2)-1)^6)/(t^2)
this is what i have so far and my computer says its not right...can anyone help me please!!!

Answer 1

f(t) = (t² - 1)^(6) / t

f `(t)

( t )(6)(t² - 1)^(5)(2t) - (t² - 1)^(6)(1)
-----------------------------------------
t²

(t² - 1)^(5) [12t² - (t² - 1)]
--------------------------------
t²

(t² - 1)^(5) (11t² + 1)
--------------------------
t²

Answer 2

f(t) =(t^2 - 1)^6 /t

f'(t) = (6(t^2 - 1)^5*2t + ((t^2 - 1)^6)/t^2
f'(t) = (12(t^2 - 1)^5)/t^2 + (t^2 - 1)^6)/t^2

Answer 3

f'(t) = ...

t (6(t^2-1)^5*2t) - (t^2 - 1)^6
----------------- -------------------- = f'(t)
t^2

12t^2 (t^2-1)^5) - (t^2 - 1)^6
______________________
t^2

(t^2-1)^5 [ 12t^2 - (t^2-1)]
___________________
t^2

(t^2-1)^5 [ 12t^2 - t^2+1]
___________________
t^2

(t^2-1)^5 [ 11^2 +1)]
----------- ---------------- = f'(t)
t^2