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

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Answer 1 · 2017-09-25T18:19:33

Given: #y=(x-y)^2/(x+y); x!=-y#

Multiply both sides by #x+y#:

#xy+y^2 = (x-y)^2#

Expand the square:

#xy+y^2 = x^2 -2xy+y^2#

Combine like terms:

#3xy = x^2#

Divide both sides by #3x#:

#y = 1/3x#

#dy/dx = 1/3#

If you do not believe the result, here is a graph of #y=(x-y)^2/(x+y)# to prove that is merely a line with a slope of #1/3#:

www.desmos.com/calculator