How do you find #dy/dx# by implicit differentiation of #(x+y)^3=x^3+y^3# and evaluate at point (-1,1)?

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Answer 1 · 2017-04-06T17:09:00

#(dy)/(dx)=-(2xy+y^2}/(x^2+2xy)# and at #(-1,1)#, #(dy)/(dx)=-1#

As #(x+y)^3=x^3+y^3#, taking differential on both sides implicitly, we get

#3(x+y)^2xx(1+(dy)/(dx))=3x^2+3y^2(dy)/(dx)#

or #3(dy)/(dx)=3{x^2-(x+y)^2}#

or #(dy)/(dx)={x^2-(x+y)^2}/{(x+y)^2-y^2}#

or #(dy)/(dx)=-(2xy+y^2}/(x^2+2xy)#

and at #(-1,1)#

#(dy)/(dx)=-(2(-1)xx1+1^2}/(1^2+2(-1)(1))=-(-1)/(-1)=-1#