How do you differentiate #2xy - y^2 = 1#?

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Answer 1 · 2015-07-15T21:59:41

#y'=-(y)/((x-y))#

#2xy-y^2=1#

#D(2xy-y^2)=D(1)#

Apply the product rule and the chain rule #rArr#

#2xy'+2y-2y.y'=0#

#2y'(x-y)=-2y#

#2y'=-(2y)/((x-y))#

#y'=-(y)/((x-y))#