Jeanette Durham
asked on
find all stationary points of f(x,y) = xe^-x (y^2-y)
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!!!!!!
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!!!!!!
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
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
ASKER
(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
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
ASKER
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
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
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
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
ASKER
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
so this would nbe
fy = 2y(xe^-x) - (xe^-x) or (xe^-x)(2y-1) ?
Zach
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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.