## anonymous 5 years ago implicit differentiation: sin((x^2)(y^2))=x

1. anonymous

ok

2. anonymous

is this sin [ x^2y^2] = x ? or (sin x^2 ) y^2 = x

3. anonymous

first one

4. anonymous

ok

5. anonymous

6. anonymous

make you fan?

7. anonymous

jkwon, let me answer it. i got this

8. anonymous

$\sin(x^2y^2)=x$ $\cos(x^2y^2)(2xy^2+x^2(2y)(y')=1$ $2xy^2+x^2(2y)y'=1/[\cos(x^2y^2)]$

9. anonymous

grrrr

10. anonymous

oh ok

11. anonymous

oh wow

12. anonymous

how do you use latex?

13. anonymous

Oh I only did half of it

14. anonymous

could you explain chain rule?

15. anonymous

sure

16. anonymous

how do you type latex like type integral x dx

17. anonymous

So you treat the problem like an onion. You do the outer layer and then you go to the second layer, then the third. So the outer layer is the cosine, and the inner layer is the x^2y^2

18. anonymous

So then....

19. anonymous

Oops i mean sine is the outer layer

20. anonymous

yeah i gotcha. so you just derive each in parts depending on parenthesis

21. anonymous

pretty much yeah ^^

22. anonymous

remember the inside is the product

23. anonymous

haha, jk i think you have this in the bud

24. anonymous

huh?

25. anonymous

nvm

26. anonymous

well im a math major, so whatever needs help, I like to solve it

27. anonymous

whats a non archimedean ordered field?

28. anonymous

can you ask that in a separate question? you're distracting him from helping me

29. anonymous

so yeah, the product rule for the x^2y^2

30. anonymous

yuppp

31. anonymous

notgoodatmath, dont speak unless told to

32. anonymous

i have 15 fans, i have solved a hundred math problems

33. anonymous

what is Y'?

34. anonymous

I just want help with this problem, and you aren't helping me. Your're doing the opposite actually.

35. sandra

hey cantorset, use the equation editor in the lower left hand side of the reply box

36. sandra

for latex

37. anonymous

oh ok did you see my comment on n^n / n!

38. sandra

oh did not check heh, will do so

39. anonymous

i dont see that as an argument what you did to show it converges