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

(1^3+2^4)

HI!!

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

partial derivatives?

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

lol yes it is. sorry

both partials are zero everywhere

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

lol now i'm confused..

ooh okay sorry about that. thanks:D

maybe i am not understanding the question correctly

is the answer 8?

can you post a screen shot or sommat?

okay just give me a second

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

are you able to make that out?

lol you forgot the exponent!!

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

LOL oh right i did, hehe SO SORRY!!

thats correct:D

How would i approach this question?

is that part clear?

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

yeah sort of..

okay cool thanks:D

what exactly are you learning?
mixed partials should be the same ordinarily

partial derivatives

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

ah!
so you need to go through the process of doing the derivative
in that case, go for it.

okay cool, thanks dude:D