How do you implicitly differentiate $- y^{2} = x y^{3} - x^{4} y$ $-y^2=xy^3-x^4y$ ?

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How do you implicitly differentiate $- y^{2} = x y^{3} - x^{4} y$ $-y^2=xy^3-x^4y$ ?

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Answer 1 · 2016-05-16T03:28:08

Given -

$- y^{2} = x y^{3} - x^{4} y$ $-y^2=xy^3-x^4y$

#-2ydy/dx=[x.3ydy/dx+y^3 . 1]-[x^4dy/dx+y.4x^3]#

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

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How do you implicitly differentiate $- y^{2} = x y^{3} - x^{4} y$ $-y^2=xy^3-x^4y$ ?

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How do you implicitly differentiate -y^2=xy^3-x^4y −y2=xy3−x4y-y^2=xy^3-x^4y ?

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How do you implicitly differentiate $- y^{2} = x y^{3} - x^{4} y$ $-y^2=xy^3-x^4y$ ?