Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

sorry, my bad!

Answer 5

\[s ^{1/3} . s ^{-2}\] = s\[s ^{-5/6}\]

Answer 6

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

Answer 7

1/s^(5/3)

Answer 8

thank you, jgeorge! Thanks so much!

Answer 9

any time :)

Answer 10

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

Answer 11

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

Answer 12

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?

Answer 13

he confused 3*2

Answer 14

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

Answer 15

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

Answer 16

yes

Answer 17

and you want to derive that?

Answer 18

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! ; )

Answer 19

no keeping secrets now lol

Answer 20

\[1+s ^{-5/3}\]

Answer 21

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

Answer 22

\[(-5/3)s^{(-5/3 - 3/3)}\]

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

does that help?

Answer 27

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!

Answer 28

its the same steps as derive \(x^4\)

Answer 29

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

Answer 30

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

Answer 31

lets take this one step at a time....

\[s^{(a)} \iff a*s^{a-1}\]
\[\frac{-5}{3}s^{(\frac{-5}{3}-\frac{3}{3})} \iff \frac{-5 s^{-8/3}}{3}\]

Answer 32

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

Answer 33

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

Answer 34

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?

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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!

Answer 41

The s^1/3 part.

Answer 42

and the posts i just did explain that :)

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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!

Answer 47

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

Answer 48

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

Answer 49

Does \(s^8\) mean: \(s.s.s.s.s.s.s.s\) ?

Answer 50

yes.

Answer 51

Does \(s.s.s.s.s.s.s.s\) equal \((s.s.s).(s.s.s).(s.s)\) ?

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

Answer 52

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

Answer 53

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

Answer 54

OH!!!!

Answer 55

and heres why:

\(\sqrt[3]{s^8} \iff \sqrt[3]{s^3s^3s^2} \iff \sqrt[3]{s^3}\sqrt[3]{s^3}\sqrt[3]{s^2}\)

Answer 56

\[\sqrt[3]{s^3} = s\]

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

Answer 57

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

Answer 58

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

Answer 59

Maybe in YOUR head, but not MINE!

Answer 60

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

Answer 61

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

Answer 62

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?

Answer 63

right.

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

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!!!!! : (

Answer 70

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

Answer 71

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

Answer 72

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).

Answer 73

:) youre doing fine :)

Answer 74

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

Answer 75

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

Answer 76

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

interactmath.com

Answer 77

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

Answer 78

Yeah, what is a sirttes anyways?

Answer 79

it my brain cells jockeying for position lol

Answer 80

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

Answer 81

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

Answer 82

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