If y = (2x- 1 / x+7 )^3. What is y' ?

y= (2x-1)/(x+7)^3

Let y=
(u/v)^3

==> y' = 3*(u/v)'
(u/v)^2

But (u/v)' =
(u'v-uv')/v^2

==> y' = 3*[(u'v-uv')/v^2]
*(u/v^2)

             = 3*[uu'v -
u^2*v]/v^4

u= 2x-1 ==> u' =
2

v= x+7 ==> v' = 1

Now
substitute:

==> y' = 3[2(2x-1)(x+7) - (2x-1)^2
]/(x+7)^4

          = 6(2x^2+13x -7) - (4x^2 -4x +
1)]/(x+7)^4

           = (12x^2 + 78x - 42 - 4x^2 + 4x
-1)/(x+7)^4

            =  (8x^2 +82x -
43)/(x+7)^4

 ==> y' = (8x^2 + 82x -
43)/(x+7)^4