Question

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

Answer 1

its impossible because the formula is wrong

Answer 2

his yes it is

Answer 3

were do you get asomthing like that lol

Answer 4

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

Answer 5

answer is 6

Answer 6

How did you get that??

Answer 7

easy you do not know

Answer 8

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

Answer 9

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

Answer 10

and i do not know how he did it

Answer 11

okay thanks

Answer 12

whats the question

Answer 13

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

Answer 14

with respect to what?

Answer 15

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

Answer 16

make me your fan and more information will come soon

Answer 17

thanks lol

Answer 18

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

Answer 19

with respect to all of them lol

Answer 20

i will do this on paper

Answer 21

thanks a lot

Answer 22

didnt post?

Answer 23

how do do a partial derivative in respect to everything?

Answer 24

nope...

Answer 25

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

Answer 26

then it is in respect to x

Answer 27

yes

Answer 28

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

Answer 29

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

Answer 30

oh. ok

Answer 31

so what is so difficult about this?

Answer 32

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

Answer 33

benito, not nice

Answer 34

Thanks cantorset!!!

Answer 35

now we find fxx, fxy, fyx, and fyy

Answer 36

whats cool, it turns out fxy = fyx always

Answer 37

i see

Answer 38

first fundamental theorem of partial derivatives