Find derivative for #(y+x)/(y-2)- (3y)/(x^2)=0# ?

1 Answer
Sonnhard
Jul 13, 2018

#y'=-(3x^2+2xy)/(x^2-6y+6)#

Explanation:

Assuming that #y=y(x)# is hold.
Multiplying the given equation by #x^2(y-2)# we get

#x^2(y+x)-3y(y-2)=0#
expanding

#x^3+x^2y-3y^2+6y=0#
Differentiating with respect to #x#

#2xy+x^2y'+3x^2-6yy'+6y'=0#

#y'(x^2-6y+6)=-3x^2-2xy#
so #y'# is given by

#y'=-(3x^2+2xy)/(x^2-6y+6)# if #x^2-6y+6ne 0#

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