How do you implicitly differentiate $11=-xy^2-(xy)/(2x-y)$ ?

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Answer 1 · 2016-11-03T16:51:28

$dy/dx=(y^3-4xy^2-y-22)/(4x^2y -3xy^2 +x-11)$

$11=-xy^2-(xy)/(2x-y)$

$11(2x-y)=-xy^2(2x-y)-xy$

$22x-11y=-2x^2y^2+xy^3-xy$

$22-11dy/dx=-4xy^2-4x^2y dy/dx+y^3+3xy^2 dy/dx-y-xdy/dx$

$4x^2y dy/dx-3xy^2 dy/dx+xdy/dx-11dy/dx=y^3-4xy^2-y-22$

$dy/dx(4x^2y -3xy^2 +x-11)=y^3-4xy^2-y-22$

$dy/dx=(y^3-4xy^2-y-22)/(4x^2y -3xy^2 +x-11)$

How do you implicitly differentiate 11=-xy^2-(xy)/(2x-y)11=-xy^2-(xy)/(2x-y)?