How do you implicitly differentiate $y - {(3 y - x)}^{2} - \frac{1}{y^{2}} = x^{3} + y^{3} - x y$ $y-(3y-x)^2-1/y^2=x^3 + y^3- xy$ ?

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Answer 1 · 2016-04-01T10:03:14

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

$y - {(3 y - x)}^{2} - \frac{1}{y^{2}} = x^{3} + y^{3} - x y$ $y-(3y-x)^2-1/y^2=x^3+y^3-xy$

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

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

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

How do you implicitly differentiate y-(3y-x)^2-1/y^2=x^3 + y^3- xyy−(3y−x)2−1y2=x3+y3−xy y-(3y-x)^2-1/y^2=x^3 + y^3- xy?