Ok, so first you have to realize that 2xy^2 is a u-v problem
So...
U:2x V:y^2
U':2 V':2y(dy/dx)
So dy/dx: 2y^2+2x2y(dy/dx)
dy/ dx = -2y^2/(2x2y)
This then is u-v'd again to get it into the second derivative form
u:-2y^2 v:2x2y(which needs a sep u-v 2 solve)
u':-4y(dy/dx) v':4y+4x(dy/dx)
So the second derivative:
-4y(dy/dx)2x2y+-2y^2(4y+4x(dy/dx))/ (2x2y)^2
Then you plug in answer you got for dy/dx in the second derivative formula, do a bunch of 'fun' algebra and you get the answer (which, by the way is:
8y^3-2y^2(4x)-2y^2(4y)/(2x2y) (which of course, if you feel up to a challege in you orginizational skills, you can simplify
Use implicit differentiation to find the second derivative:
2xy^2=4 (nothing is in parenthesis)
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