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what is partial derivative of f(x,y) = x^3 - 3xy^2 + 6y^2
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fsubx=3x^2-3y^2 fsuby=-6xy+12y
Hi, do you need to find it in both terms of x and y?
In terms of x: 3x^2 - 3y^2 + 0 In terms of y: 0 -6xy + 12y
yes thanks.. let me take a look and see if I understand
I am trying to find the local extrema
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I am having a hard time solving x and y to find the critical points
can you show me how
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