How do you find the second partial derivative of f(x,y)=cos^2^xsin^2^y?

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How do you find the second partial derivative of f(x,y)=cos^2^xsin^2^y?

Mathematics
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its impossible because the formula is wrong
his yes it is
were do you get asomthing like that lol

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Other answers:

it's f(x,y)=cos(squared)xsin(squared)y
answer is 6
How did you get that??
easy you do not know
no that's why i'm asking... lol
i can not tell you cuz ther is diff way to do it
and i do not know how he did it
okay thanks
whats the question
How do you find the second partial derivative of f(x,y)=cos(squared)xsin(squared)y?
with respect to what?
we start with fx, and fy, then we get fxx, fxy, fyx and fyy, notice that fxy = fyx
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thanks lol
ok with respect to what so you want all the partial derivatives? ok one sec
with respect to all of them lol
i will do this on paper
thanks a lot
didnt post?
how do do a partial derivative in respect to everything?
nope...
fx = sin^2 y * 2 cos x (-sin x), treat y as a constant
then it is in respect to x
yes
thats first partial wrt to x , wrt means with respect to
fy = cos^2 x * 2 sin y cos y (treat x as a constant)
oh. ok
so what is so difficult about this?
I didn't know how to do it obviously...
benito, not nice
Thanks cantorset!!!
now we find fxx, fxy, fyx, and fyy
whats cool, it turns out fxy = fyx always
i see
first fundamental theorem of partial derivatives

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