How do you implicitly differentiate #-y= x^3y^2-x^2y^3-2xy^4 #?

1 Answer

#y'=(3x^2y^2-2xy^3-2y^4)/(3x^2y^2+8xy^3-2x^3y-1)#

Explanation:

From the given #-y=x^3y^2-x^2y^3-2xy^4#
differentiate both sides of the equation with respect to x

#d/dx(-y)=d/dx(x^3y^2-x^2y^3-2xy^4)#

#-y'=x^3*d/dx(y^2)+y^2*d/dx(x^3)-[x^2*d/dx(y^3)+y^3*d/dx(x^2)]-2*[x*d/dx(y^4)+y^4*d/dx(x)]#

#-y'=x^3*2yy'+3x^2y^2-3x^2y^2y'-2xy^3-8xy^3y'-2y^4#

Isolate all terms with y' in one side of the equation

#-y'-2x^3yy'+3x^2y^2y'+8xy^3y'=3x^2y^2-2xy^3-2y^4#

Factor out the common monomial factor #y'#

#(-1-2x^3y+3x^2y^2+8xy^3)y'=3x^2y^2-2xy^3-2y^4#

Divide both sides of the equation by the coefficient of #y'# which is #(-1-2x^3y+3x^2y^2+8xy^3)#

#((-1-2x^3y+3x^2y^2+8xy^3)y')/((-1-2x^3y+3x^2y^2+8xy^3))=(3x^2y^2-2xy^3-2y^4)/(-1-2x^3y+3x^2y^2+8xy^3)#

#y'=(3x^2y^2-2xy^3-2y^4)/(3x^2y^2+8xy^3-2x^3y-1)#

God bless....I hope the explanation is useful.