Evaluate the derivative of y^3 where y = x/(x^3+3)^1/3

Y = x/(x^3+3)^(1/3)

To find
the derivative of y^3 with respect to
x.

Solution:

We know d/dx(y^3)
= 3y^2* dy/dx.............(1).

dy/dx = y' ={
x/[(x^3+3)^(1/3)]}' = (u/v)' form

(u/v) ' = {u'*v
+u*v'}/v^2.

u = x, u' = 1

v=
{(x^3+3)^(1/3)}' , v' = {(x^3+3)^(1/3)}' = (1/3){x^3+3) ^(1/3-1) *
(x^3+3)'

v' = (1/3)(x^3+3)^(-2/3) *
(3x^2)

v' =
x^2/(x^3+3)^(2/3).

Therfore y '= (u/v)' = {1 (x^3+3)^(1/3)
+ x^2/(x^3+3)}/(x^3+3)^(2/3)  = {(x^3+3)
+x^2}/(x^3+3)^(4/3)

y' =
(x^3+x^2+3)/(x^3+3)^(4/3)

We  substitute this value of the
value of y' in (1) and we get:

dy^3/dx = 3 y^2 {
(x^3+x^2+3)/(x^3+3)^(4/3)

d/dx(y^3) =  3 {x/x^3+3)^(1/3)}^2
{(x^3+x^2+3)/(x^2+3)^(4/3)

= 3x^2(x^3+x^2+3)/
(x^3+3)^(6/3)

d/dx(y^3) =
3x^2(x^2+3)/(x^3+3)^2.