At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

its impossible because the formula is wrong

his yes it is

were do you get asomthing like that lol

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

make me your fan and more information will come soon

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