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find all stationary points of f(x,y) = xe^-x (y^2-y)

Posted on 2011-03-02
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I have the above question
f(x,y) = xe^-x(y^2-y)

I need to find the following:

fx = e^-x(y^2-y) + xe^-x(y^2-y)(-1) is as far as I got one this one...
fy
fxx
fyy
fxy

I'm stuck on how to differentiate this.  The chain rule is a mess on this!  Any help you can provide would be GREATLY appreciated!!!!!!
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Question by:JeffreyDurham
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by:phoffric
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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.
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by:deighton
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is that

xe^-(x (y^2-y))

or

(xe^-x) (y^2-y)

your answer implies it is the second of those, in which case you are on the right track


finding partial deriviatives is a case of using the chain or product rule here, with x or y treated as a constant if not differentiating by that one
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by:JeffreyDurham
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(xe^-x) (y^2-y)

is correct...

the e is what is causing me trouble... i am aware of the chain rule, this is where I go to for fy but I seem to get really confused here...

f'x(g(gx))+g'x(f(x))

f(x) = (xe^-x)(y^2-y) (whole term?)
g(x) = y^2-2

f'x = 0?  no y term by itself?  this is where I am unclear
g'x = 2y

thusly... fyy = 0*2y+ (xe^-x)(y^2-y)*2x ?  

I understand this would break down further
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Author Comment

by:JeffreyDurham
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let me sort of clarify my confusion because examing what I did this is where I am having trouble.

in the fy, what should I be using as the f(x) and g(x).  From what i see I have 3 seperate terms.  

f(x) candidate one:  whole term, there is no y variable so the derivitive of y would be 0.
f(x) candidate two:  xe^-x
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by:phoffric
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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.

For derivatives of e w.r.t. chain rule, see: http://rdsrc.us/OfpVSC
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Author Comment

by:JeffreyDurham
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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) ?



Zach
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Accepted Solution

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phoffric earned 500 total points
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Right!
Using k (to keep things less ugly):

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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