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

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

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Answer 1 · 2016-11-03T16:51:28

$\frac{d y}{d x} = \frac{y^{3} - 4 x y^{2} - y - 22}{4 x^{2} y - 3 x y^{2} + x - 11}$ $dy/dx=(y^3-4xy^2-y-22)/(4x^2y -3xy^2 +x-11)$

Explanation:

$11 = - x y^{2} - \frac{x y}{2 x - y}$ $11=-xy^2-(xy)/(2x-y)$

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

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

$22 - 11 \frac{d y}{d x} = - 4 x y^{2} - 4 x^{2} y \frac{d y}{d x} + y^{3} + 3 x y^{2} \frac{d y}{d x} - y - x \frac{d y}{d x}$ $22-11dy/dx=-4xy^2-4x^2y dy/dx+y^3+3xy^2 dy/dx-y-xdy/dx$

$4 x^{2} y \frac{d y}{d x} - 3 x y^{2} \frac{d y}{d x} + x \frac{d y}{d x} - 11 \frac{d y}{d x} = y^{3} - 4 x y^{2} - y - 22$ $4x^2y dy/dx-3xy^2 dy/dx+xdy/dx-11dy/dx=y^3-4xy^2-y-22$

$\frac{d y}{d x} (4 x^{2} y - 3 x y^{2} + x - 11) = y^{3} - 4 x y^{2} - y - 22$ $dy/dx(4x^2y -3xy^2 +x-11)=y^3-4xy^2-y-22$

$\frac{d y}{d x} = \frac{y^{3} - 4 x y^{2} - y - 22}{4 x^{2} y - 3 x y^{2} + x - 11}$ $dy/dx=(y^3-4xy^2-y-22)/(4x^2y -3xy^2 +x-11)$

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

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How do you implicitly differentiate 11=-xy^2-(xy)/(2x-y)11=−xy2−xy2x−y11=-xy^2-(xy)/(2x-y)?

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