How do you find #(dy)/(dx)# given #x^3+y^3=3xy^2#?

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Answer 1 · 2016-11-17T09:26:14

#(dy)/(dx)=((y+x)(y-x))/(y(y-2x))#

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

the RHS will need the product rule.

#3x^2+3y^2(dy)/(dx)=3y^2+6xy(dy)/(dx)#

rearrange for# (dy)/(dx)#, and simplify the algebra.

#3y^2(dy)/(dx)-6xy(dy)/(dx)=3y^2-3x^2#

#(dy)/(dx)(3y^2-6xy)=3(y^2-x^2)#

#(dy)/(dx)=(3(y+x)(y-x))/(3y(y-2x))#

#(dy)/(dx)=((y+x)(y-x))/(y(y-2x))#