How do you differentiate #2x^2-3xy+3y^2=2# at (1,1)?

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Answer 1 · 2015-05-02T09:02:45

To do this you differentiate implicitly

#2x^2-3xy+3y^2=2# becomes,

#4x-3y -3x(dy)/(dx)+6y(dy)/(dx)=0#

#=>(6y-3x)(dy)/(dx)=3y-4x#

#=>(dy)/(dx)=(3y-4x)/(6y-3x)#

At point #(1,1)# # y=1 and x=1#

#=> (dy)/(dx)=(3-4)/(6-3)=-1/3#