How do you implicitly differentiate # y-(3y-x)^2-1/y^2=x^3 + y^3- xy#?

1 Answer
Apr 1, 2016

By differentiating #x# and #y# like your usual differetiation practices.

Explanation:

#y-(3y-x)^2-1/y^2=x^3+y^3-xy#

I will let #(dy)/(dx)=y'#

Now begin the implicit differentiation

#y'-2(3y-x)(3y'-1)-(-2)1/y^3y'=3x^2+3y^2y'-(y+xy')#

I moved all the terms to one side and factored out #y'#

#y'[7-18y+2/y^2-3y^2+x]+7y-2x-3x^2=0#

Trick is; after differentiating any form of #y#, add #y'#

Example: #d/dxy^2=2yy'#

Hope this helps. Cheers