Thread: Find y'' by implicit differentiation help!

x^3 + y^3 =6xy

No need for d/dx just dy/dx for when you find the derivative of y only.

My steps:

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

dy/dy = (3x^2-6y) /(6x-3y^2)
= (x^2-2y) / (2x-y^2)

y'' = [ (2x- 2(dy/dx)) (2x-y^2) ] - [(x^2-2y)(2-2y(dy/dx))]
= [ 2x-2(x^2-2y/2x-y^2) ] - [(x^2-2y)(2-2y(x^2-2y/2x-y^2)) ] / (2x-y^2)^2

If your having trouble understanding, when i find the derivative of y, its the derivative and dy/dx, where i replace dy/dx in y'' with the dy/dx from y' .

I dont know how to actually put symbols i cant find any button to place actually division signs and such so i clear this up a bit


So then i simplified the complex fraction by multiplying by 2x-y^2

[ 2x(2x-y^2)-2(x^2-2y) ] - (x^2-2y)(2x-y^2) (2(2x-y^2)) (2y(x^2-2y))
4x^2-2y^2 -2x^2-4y - 4y(x^2-2y)^2(2x-y^2)^2
2x^2-2xy^2-4y -4y(x^2-2y)^2(2x-y^2)^2 / (2x-y^2)^2


Is any of this right, that final shown above, not final but mid way, seems way too complicated and much work for such a simple question. All that factoring and such seems alot.


If clarification is needed dont hesitate

OK, for starters, it helps if you don't switch between Leibnitz and Newtonian notation. Stick to one or the other. I tend to prefer Leibnitz.

Second, you need to understand that is an OPERATOR meaning "take the derivative with respect to x", while is a function, namely the derivative of y with respect to x. It can also be written as y'(x) if you prefer Newtonian notation.

The basic idea is that y is a function of x, so if you have a function of y, you can differentiate it with respect to x.



Now take the derivative of both sides again to find the second derivative.