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July 30, 2014

July 30, 2014

Posted by **Liz** on Tuesday, December 14, 2010 at 10:15pm.

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?

- 12th Grade Calculus -
**Reiny**, Tuesday, December 14, 2010 at 10:39pmsimplify 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.

- 12th Grade Calculus -
**mulugeta**, Thursday, June 5, 2014 at 4:55amggggggggggggggggggggg

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