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June 26, 2016
Posted by **Anonymous** on Thursday, March 1, 2012 at 11:37pm.

I got the first derivative which was x^3/y^3 and no im am stuck on using implicit differentiation for the second derivative part

- calculus -
**Steve**, Friday, March 2, 2012 at 2:18pmy^4 - x^4 = 16

4y^3 y' - 4x^3 = 0

y' = x^3/y^3

y'' = (3x^2*y^3 - x^3*3y^2*y')/y^6

= 3x^2(y-xy')/y^4

= 3x^2 (y-x*x^3/y^3)/y^4

= 3x^2 (y^4 - x^4)/y^7

= 48x^2/y^7

or,

y^3 y' = x^3

3y^2 y'^2 + y^3 y'' = 3x^2

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

y^3 y'' = 3x^2 - 3x^6/y^4

y'' = 3x^2 (y^4 - x^4)/y^7