How do you use implicit differentiation to find dy/dx given xy^2+x^2y=xxy2+x2y=x?

1 Answer
Nov 5, 2016

dy/dx = (1 - 2xy - y^2)/(2xy + x^2)dydx=12xyy22xy+x2

Explanation:

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)ddx(u+v)=dudx+dvdx

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)