Define y=x^(1/3) and implicitly derive y^3 = x?

Question

By defining y = x^(1/3) and using implicit differentiation on the identity y^3 = x, derive:

d/dx [x^(1/3)] = (1/3)x^(-2/3).

Please show how you worked it out, thanks very much guys

Answer 1

y^3 = x

implicitly differentiating both sides w.r.t x and using the chain rule [ y^3 is a function of y and y is a function of x ]

3y^2 * y' = 1

y' = 1 / 3y^2

  = 1 / 3x^(2/3)   [ y = x^1/3 ]

  = 1/3 * x^(-2/3).

Answer 2

y^3 = x

=> 3y^2 dy/dx = 1

=> dy/dx = (1/3) y^(-2)

=> d/dx [x^(/3)] = (1/3) (y^3)^(-2/3)

= (1/3) x^(-2/3).