I think you should first ask questions on chain rule before attempting partial derivatives (and/or vice versa). And then you should show what you know. If you would like practice problems on the chain rule and partial derivatives, you can go here: http://cow.temple.edu/~cow/cgi-bin/manager
and open the index and find practice problems "Chain Rule" and "Partial derivatives". If you get stuck on a problem, then you can ask separate questions on it until you understand both chain rule and partials.
BTW - since in fx, e^-x(y^2-y) is a common expression in both terms, why not factor it out.
fy is partial derivative of f(x,y) = partial/dy { f(x,y) } = partial/dy { (xe^-x) (y^2-y) }
then x is treated as a constant, which means that (xe^-x) is a constant; call it K = (xe^-x)
so, fy = partial/dy { K (y^2-y) }
since now only y is showing, take straight derivative:
fy = d/dy { K (y^2-y) } = d/dy { K y^2 - Ky) }
When you complete d/dy calculation, plug back in (xe^-x) for K.
ah that is helpful. where i have been messing up is leaving the xe^-x in for the fy. it was getting extremely confusing. So it was not necessary for me to chain rule this part since the k is a constant?
so this would nbe
fy = 2y(xe^-x) - (xe^-x) or (xe^-x)(2y-1) ?
for partial w.r.t. y: f(x,y) = k(y^2-y) where k = xe^-x
fy = k2y - k = k( 2y - 1 )
= (xe^-x)( 2y - 1 )
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http://cow.temple.edu/~cow/cgi-bin/manager
and open the index and find practice problems "Chain Rule" and "Partial derivatives". If you get stuck on a problem, then you can ask separate questions on it until you understand both chain rule and partials.
BTW - since in fx, e^-x(y^2-y) is a common expression in both terms, why not factor it out.