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

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### 2 Answers

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**

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

We use y' = ( u(x)/v(x) )' = {u'(x)v(x)-u(x)v'x)}/(v(x))^2,

u(x) =2x-1.

u'(x) = (2x-1)' = 2

v(x) = (x+7)^3

v'(x) = {(x+7)^3}' = 3(x+7)^2,.

Therefore f'(x) = {2*(x+7)^3 - (2x-1)*3(x+7)^2}/(x+7)^6

f'(x) = {2(x+7) -3(2x-1)(x+7)}/(x+7)^4

f'(x) = { 2x+14- 6x^2 -39x+21}/(x+7)^6

f'(x) ={-6x^2-37x+35)/(x+7)^2

f'(x) = -{6x^2+37x+35)/(x+7)^4