Question

hey guys, some help please..
if f(x,y) = (x^3+y^4) then f(xy)-f(yx) at the point (1,2) is.. how would i solve this question. Thanks in advance:D

Answer 1

you would substitute 1 in for x and 2 in for y

Answer 2

(1^3+2^4)

Answer 3

HI!!

Answer 4

is it \[\huge f_{xy}-f_{yx}\]?

Answer 5

partial derivatives?

Answer 6

no that can't be right, both second partials are zero
must be something else

Answer 7

lol yes it is. sorry

Answer 8

both partials are zero everywhere

Answer 9

\[f(x,y) = (x^3+y^4) \]\]
\[f_x(x,y)=3x^2\] so \[f_{xy}(x,y)=0\]

Answer 10

lol now i'm confused..

Answer 11

the partial wrt x, treat y as a constant so it is just \(3x^2\)
then if you take the partial wrt y, there is no y in it, so it is just zero

Answer 12

ooh okay sorry about that. thanks:D

Answer 13

maybe i am not understanding the question correctly

Answer 14

is the answer 8?

Answer 15

its a multiple choice question. i'm not really sure how to approach the question. could it be that the question is incorrect?

Answer 16

can you post a screen shot or sommat?

Answer 17

okay just give me a second

Answer 18

my pc is a little bit slow this evening, please bear with me

Answer 19

are you able to make that out?

Answer 20

lol you forgot the exponent!!

Answer 21

\[f(x,y) = (x^3+y^4)^5\]

Answer 22

LOL oh right i did, hehe SO SORRY!!

Answer 23

thats correct:D

Answer 24

How would i approach this question?

Answer 25

i think it is still has to be zero, but at now we can compute it
first
\[f_x=5(x^3+y^3)^4\times 3x^2\] via the chain rule

Answer 26

is that part clear?

Answer 27

oops gotta run, you will get more help i am sure

Answer 28

yeah sort of..

Answer 29

okay cool thanks:D

Answer 30

what exactly are you learning?

mixed partials should be the same ordinarily

Answer 31

partial derivatives

Answer 32

its a new section of work so im really not too familiar with it

Answer 33

ah!

so you need to go through the process of doing the derivative

in that case, go for it.

Answer 34

okay cool, thanks dude:D