anonymous 5 years ago ok, here's another exponent problem! If you guys don't mind...cube root of s over s^2 = ?

1. anonymous

$\sqrt[3]{s}\div s ^{2}$

2. mathteacher1729

I believed this was asked just a little while ago today...

3. anonymous

why is to the -1/3 and not positive 1/3?

4. anonymous

5. anonymous

$s ^{1/3} . s ^{-2}$ = s$s ^{-5/6}$

6. amistre64

${s^{1/3} \over s^2} = {1 \over s^{2-\frac{1}{3}}}$

7. amistre64

1/s^(5/3)

8. anonymous

thank you, jgeorge! Thanks so much!

9. anonymous

any time :)

10. anonymous

Amistre64, where in the heck have you been? I have calculus problems to solve, you know!!! ; )

11. amistre64

i had to do laundry; and air up my tire; put oil in the car... you know, nonmathical stuff lol

12. anonymous

Well, next time do those things when I DON'T need your help! Seriously, how have you been? Can you help me now with the antiderivative of that exponent you gave me? And wait a minute, why is it to the -5//6 and not -5/3?

13. amistre64

he confused 3*2

14. amistre64

2-(1/3) = (3*2 -1)/3

15. anonymous

so it is s^-5/3, then? right?

16. amistre64

yes

17. amistre64

and you want to derive that?

18. anonymous

Need help finding the antiderivative of this if you don't mind then. Let me try to attach an equation just to look professional, like i know what I'm doing! ; )

19. amistre64

no keeping secrets now lol

20. anonymous

$1+s ^{-5/3}$

21. anonymous

Need to find the antid of that! I know it's s-something. but it's the "something" i get stuck on, sort of!

22. amistre64

$(-5/3)s^{(-5/3 - 3/3)}$

23. amistre64

$\frac{-5}{3 s^2 \sqrt[3]{s^2}}$

24. amistre64

$$*\sqrt[3]{s}$$ top and bottom to rationalise the denm

25. amistre64

$\frac{-5 \sqrt[3]{s}}{3s^3}$

26. amistre64

does that help?

27. anonymous

Well I am trying to figure out the first step where had -5 over all that gobbledy-gook. I don't get that part at all!

28. amistre64

its the same steps as derive $$x^4$$

29. amistre64

$$\ 'exp' * x^{(exp-1)}$$

30. anonymous

Where did the s^2 come fronm in the denominator? I see about the cube root, but not that.

31. amistre64

lets take this one step at a time.... $s^{(a)} \iff a*s^{a-1}$ $\frac{-5}{3}s^{(\frac{-5}{3}-\frac{3}{3})} \iff \frac{-5 s^{-8/3}}{3}$

32. amistre64

$\frac{-5}{3s^{8/3}} \iff \frac{-5}{3s^{6/3}s^{2/3}}$

33. amistre64

$\frac{-5}{3s^2 s^{2/3}}*\frac{s^{1/3}}{s^{1/3}}=\frac{-5s^(1/3)}{3s^2 s}$

34. anonymous

Ok, I see where you are getting those numbers, but why do you have to do the 6/3 2/3 thing in the denominator in the first place? Why do you have to break it up like that? You can't just leave it? Why 2 "s" 's?

35. amistre64

$\frac{-5 \sqrt[3]{s}}{3s^3}$

36. amistre64

for the same reason why we dont leave $$\sqrt{36}$$ as it is :)

37. anonymous

ok, but why the s^3? I though it was s^2 in the denominator! aaarrrgh!

38. amistre64

$\sqrt[3]{s.s.s.s.s.s.s.s} \iff s^2 \sqrt[3]{s^2}$

39. amistre64

$\sqrt[3]{s^3 s^3 s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}$ $s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2}$

40. anonymous

ok, go back up if you would to where I said,"ok, I see where you get those numbers, but why do you have to do the 6/3 2/3 thing in the denominator." See that post of mine? the one right before it is yours. THATS what I don't understand!

41. anonymous

The s^1/3 part.

42. amistre64

and the posts i just did explain that :)

43. anonymous

ok, then I am just an idiot, because I don't get it at all. What to do?

44. amistre64

$$\frac{-5}{3s^{8/3}}$$ is what we get in the denominator right?

45. amistre64

your question is about the s^(8/3) and how we play with it

46. anonymous

Is that 3s^8/3? It's quite small! But yes that is my question. How to play with it. But i have other things I would rather play with. Exponents AINT one of em!

47. amistre64

Does $$s^{8/3}$$ equal $$\sqrt[3]{s^8}$$?

48. anonymous

yes, i would have to say it does. ok, go on... please!

49. amistre64

Does $$s^8$$ mean: $$s.s.s.s.s.s.s.s$$ ?

50. anonymous

yes.

51. amistre64

Does $$s.s.s.s.s.s.s.s$$ equal $$(s.s.s).(s.s.s).(s.s)$$ ? Does $$(s.s.s).(s.s.s).(s.s) \iff s^3 s^3 s^2$$ ?

52. anonymous

yes, but is there a reason you split it up like that and not like (s.s.s.s).(s.s).(s.s)?

53. amistre64

yes theres; our radical is a $$\sqrt[3]{...}$$ which means we want to group these into $$s^3$$ .

54. anonymous

OH!!!!

55. amistre64

and heres why: $$\sqrt[3]{s^8} \iff \sqrt[3]{s^3s^3s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}$$

56. amistre64

$\sqrt[3]{s^3} = s$ $s.s.\sqrt[3]{s^2} \iff s^2 \sqrt[3]{s^2 }$

57. anonymous

OMGosh! Ok, so back to the problem...I don't remember where we were even!

58. amistre64

these are the basic steps that you tend to do in your head to avoid all the work of reproving them time and again lol

59. anonymous

60. amistre64

yes, it does help to actually step thru them to gain insight ;)

61. amistre64

ok so we are left with: $\frac{-5}{3s^2\sqrt[3]{s^2}}$ and thats fine but, we maybe want to get rid of the radical in the bottom part

62. amistre64

we 'know' $$\sqrt[3]{s^3}=s$$ so we want to multily top and bttom by $$\sqrt[3]{s}$$ to get rid of it in the bottom right?

63. anonymous

right.

64. amistre64

$\frac{-5 \sqrt[3]{s}}{3s^3}$

65. anonymous

wait, if you multiply the numerator by that radical and that radical =s, why do you still have the radical sign there?

66. amistre64

$-5 * \sqrt[3]{s} = -5\sqrt[3]{s}$

67. anonymous

yeah, but it's the cube root of s^3. That just equals s I thought?

68. amistre64

$3s^2\sqrt[3]{s^2}*\sqrt[3]{s}\iff3s^2\sqrt[3]{s^2.s}\iff3s^2\sqrt[3]{s^3}\iff 3s^2.s$ $\implies 3s^3$

69. anonymous

Are they all this hard? How in the heck do they expect newcomers here to calculus to know all this crap? I really don't understand! It's so hard!!!!! : (

70. amistre64

they expect you to have learned this in college algebra; since thats all it is

71. amistre64

if they have to reteach you what you should have already learned... then it gets rather ...... misses the point of dividing mathinto teachable sections lol

72. anonymous

BTW your answers are always the best and easiest to work with! As are you! Thank you so much! I am going to go and try towrok this problem...will you be around? Oh, and the college algebra thing? I got a LOW C! I couldn't do it then; I aced calc in college 20 years ago, and now don't know a thing, but could help my daughter with high school algebra II like a pro (sort of).

73. amistre64

:) youre doing fine :)

74. anonymous

I get the point! Do you know of a good website with the exponent rules on it?

75. anonymous

Actually, somehow in calc I'm getting a 97%!

76. amistre64

my sirttes are all books; but i know a good practie site for math: interactmath.com

77. amistre64

sirttes eh.... my uiniversal translator is on the fritz i think lol

78. anonymous

Yeah, what is a sirttes anyways?

79. amistre64

it my brain cells jockeying for position lol

80. anonymous

Will you be around for some of today? Can I visit you here again if I need to?

81. amistre64

ill be here as long as the weather holds up; im under the veranda since the libraries are closed

82. anonymous

Ok, then again...thanks from the bottom of my heart for all your wonderful help! You're the best! Giving me all that time and patience...you're great, thanks