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

- anonymous

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

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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 this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

its impossible because the formula is wrong

- anonymous

his yes it is

- anonymous

were do you get asomthing like that lol

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

it's f(x,y)=cos(squared)xsin(squared)y

- anonymous

answer is 6

- anonymous

How did you get that??

- anonymous

easy you do not know

- anonymous

no that's why i'm asking... lol

- anonymous

i can not tell you cuz ther is diff way to do it

- anonymous

and i do not know how he did it

- anonymous

okay thanks

- anonymous

whats the question

- anonymous

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

- anonymous

with respect to what?

- anonymous

we start with fx, and fy, then we get fxx, fxy, fyx and fyy,
notice that fxy = fyx

- anonymous

make me your fan and more information will come soon

- anonymous

thanks lol

- anonymous

ok with respect to what
so you want all the partial derivatives?
ok one sec

- anonymous

with respect to all of them lol

- anonymous

i will do this on paper

- anonymous

thanks a lot

- anonymous

didnt post?

- anonymous

how do do a partial derivative in respect to everything?

- anonymous

nope...

- anonymous

fx = sin^2 y * 2 cos x (-sin x), treat y as a constant

- anonymous

then it is in respect to x

- anonymous

yes

- anonymous

thats first partial wrt to x , wrt means with respect to

- anonymous

fy = cos^2 x * 2 sin y cos y (treat x as a constant)

- anonymous

oh. ok

- anonymous

so what is so difficult about this?

- anonymous

I didn't know how to do it obviously...

- anonymous

benito, not nice

- anonymous

Thanks cantorset!!!

- anonymous

now we find fxx, fxy, fyx, and fyy

- anonymous

whats cool, it turns out fxy = fyx always

- anonymous

i see

- anonymous

first fundamental theorem of partial derivatives

Looking for something else?

Not the answer you are looking for? Search for more explanations.