Question

Find d^2y/dx^2 by implicit differentiation.

x^(1/3) + y^(1/3) = 4

I know that first you must find the 1st derivative & for y prime I got 1/3x^(-2/3) + 1/3y^(-2/3) dy/dx = 0

Then for dy/dx I got
dy/dx = [-1/3x^(-2/3)] / [1/3y^(-2/3)]

I think that from here I would use the quotient rule to find the second derivative?

Answer 1

simplify your first derivative before going further
notice you can divide each term by 1/3 to get

x^(-2/3) + y^(-2/3) dy/dx = 0
dy/dx = -x^(-2/3) / y^(-2/3)
= - y^(2/3)/x^(2/3)
= - (y/x)^(2/3)

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

replace the dy/dx in the square bracket by (y/x)^2/3) and see what you get.

Still messy but a small improvement.