How do you use implicit differentiation to find dy/dx given $xy^2+x^2y=x$ ?

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Answer 1 · 2016-11-05T20:00:19

$dy/dx = (1 - 2xy - y^2)/(2xy + x^2)$

The differentiation of the given expression is determined by using differentiation of the sum and the product differentiation.

Differentiation of the sum:

$color(blue)(d/dx(u+v)=(du)/dx+(dv)/dx)$

Product differentiation:

$color(brown)((uv)'=u'v+v'u)$

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$xy^2+x^2y=x$

$rArrd/dx(xy^2+x^2y)=(dx)/dx$

$rArrcolor(blue)((d(xy^2))/dx+(d(x^2y))/dx)=1$

$rArrcolor(brown)((y^2xxdx/dx+x xx(dy^2)/dx))+color(brown)((yxxdx^2/dx+x^2xxdy/dx)=1$

$rArry^2 + 2xydy/dx +2xy + x^2dy/dx = 1$

$rArr2xydy/dx + x^2dy/dx +2xy +y^2 = 1$

$rArrdy/dx(2xy + x^2) +2xy +y^2 = 1$

$rArrdy/dx(2xy + x^2) = 1 - 2xy - y^2$

$rArrdy/dx = (1 - 2xy - y^2)/(2xy + x^2)$

How do you use implicit differentiation to find dy/dx given xy^2+x^2y=xxy^2+x^2y=x?