How do you differentiate #-2=y/(x-y^2)-y+x^2#?

1 Answer
Nov 6, 2016

#\frac{2xy^2+y-3x^2-2}{3y^2-4y-2yx^2+1-x}=\fracdydx#

Explanation:

First, you remove the fraction by multiplying by #x-y^2#
#-2(x-y^2)=y-y(x-y^2)+x^2(x-y^2)#
expand
#-2x+2y^2=y-yx+y^3+x^3-x^2y^2#
Next, I like to move all terms onto one side and put them in a good order:
#0=y^3-2y^2-x^2y^2+y-yx+x^3+2x#
Next, take the derivative:
#0=\fracddx[y^3]-\fracddx[2y^2]-\fracddx[x^2y^2]+\fracddx[y]-\fracddx[yx]+\fracddx[x^3]+\fracddx[2x]#
Use the product rule, chain rule, and power rule:
#0=3y^2\fracdydx-4y\fracdydx-(x^2\cdot2y\fracdydx+2xy^2)+\fracdydx-(y+x\fracdydx)+3x^2+2#

group all #\fracdydx# terms
#0=3y^2\fracdydx-4y\fracdydx-2yx^2\fracdydx+\fracdydx-x\fracdydx-2xy^2-y+3x^2+2#

factor #\fracdydx#
#0=\fracdydx(3y^2-4y-2yx^2+1-x)-2xy^2-y+3x^2+2#

subtract non #\fracdydx# terms
#2xy^2+y-3x^2-2=\fracdydx(3y^2-4y-2yx^2+1-x)#

divide both sides by#3y^2-4y-2yx^2+1-x#
#\frac{2xy^2+y-3x^2-2}{3y^2-4y-2yx^2+1-x}=\fracdydx#