## anonymous one year ago 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

1. ayee_ciera

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

2. ayee_ciera

(1^3+2^4)

3. misty1212

HI!!

4. misty1212

is it $\huge f_{xy}-f_{yx}$?

5. misty1212

partial derivatives?

6. misty1212

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

7. anonymous

lol yes it is. sorry

8. misty1212

both partials are zero everywhere

9. misty1212

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

10. anonymous

lol now i'm confused..

11. misty1212

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

12. anonymous

ooh okay sorry about that. thanks:D

13. misty1212

maybe i am not understanding the question correctly

14. IrishBoy123

15. anonymous

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

16. misty1212

can you post a screen shot or sommat?

17. anonymous

okay just give me a second

18. anonymous

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

19. anonymous

20. anonymous

are you able to make that out?

21. misty1212

lol you forgot the exponent!!

22. misty1212

$f(x,y) = (x^3+y^4)^5$

23. anonymous

LOL oh right i did, hehe SO SORRY!!

24. anonymous

thats correct:D

25. anonymous

How would i approach this question?

26. misty1212

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

27. misty1212

is that part clear?

28. misty1212

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

29. anonymous

yeah sort of..

30. anonymous

okay cool thanks:D

31. IrishBoy123

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

32. anonymous

partial derivatives

33. anonymous

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

34. IrishBoy123

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

35. anonymous

okay cool, thanks dude:D