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

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

Let us cube both sides:

y^3 = x^3 / *x^3 + 3)

Ley y^3 = f(x) = x^3/(x^3 + 3)= u/v

=> u = x^3 ==> u' = 3x^2

=> v= x^3 + 3 ==> v' = 3x^2

dy^3 = f'(x) = (u'v- uv')/v^2

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

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

= 9x^2/(x^3 + 3)^2

To evaluate the derivative of f^3, we'll have to determine the expression of f^3.

f^3 = [x/(x^3 +3)^1/3]^3

f^3 = x^3/(x^3 +3)

Now, we'll calculate the first derivative using the quotient rule:

(u/v)' = (u'*v-u*v')/v^2

We'll put u=x^3 and v=x^3 +3

u' = 3x^2

v' = 3x^2

(f^3)' = [3x^2*(x^3 +3) - x^3*3x^2]/(x^3 +3)^2

We'll factorize the numerator, by 3x^2:

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

We'll eliminate like terms, from numerator, and we'll remove the brackets:

**(f^3)' = 9x^2/(x^3 +3)^2**

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

Taking the cube of both the sides: y^3= x^3/(x^3+3)

Now if f(x)=g(x)/h(x), f'(x)= [g'(x)h(x)-g(x)h'(x)]/[h(x)]^2

Also for f(x)=x^n, f'(x)=n*x^(n-1)

Now we have to find the derivative of y^3 which can be taken as f(x).

g(x)=x^3 and h(x)=x^3+3

g'(x)=3x^2 and h'(x)=3x^2

Therefore f'(x)= [g'(x)h(x)-g(x)h'(x)]/[h(x)]^2

=[3x^2*(x^3+3)-(x^3)*3x^3]/(x^3+3)^2

=3x^2*(x^3+3-x^3)/(x^3+3)^2

=9x^2/(x^3+3)^2

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.