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Mathematics
OpenStudy (anonymous):

differentiate: s^1/3 - 1 / s^2/3 -1

7 years ago
OpenStudy (anonymous):

(1/3) s^-2/3 + (2/3) s^(-5/3)

7 years ago
OpenStudy (anonymous):

shouldn't you apply quotient rule rather than product rule? and for that matter, wouldn't the last bit be s^-1/3

7 years ago
OpenStudy (anonymous):

1/s^2/3 = s^(-2/3)

7 years ago
OpenStudy (anonymous):

-2/3-1 = -5/3

7 years ago
OpenStudy (anonymous):

oh okay. but what about the rule? it should be quotient.

7 years ago
OpenStudy (anonymous):

it doesnt matter. it is easier my way

7 years ago
OpenStudy (anonymous):

I know, I tried that too. but then that's not the answer :/ somehow the answer is 1/3 (s^1/3 +1)^-2 s^-2/3

7 years ago
OpenStudy (anonymous):

wait. the question is ambiguous. can you post your question using the equation button below?

7 years ago
OpenStudy (anonymous):

type in frac{1}{s} to get 1/s

7 years ago
OpenStudy (anonymous):

\[g(s) = s ^{1/3} - 1 \div s ^{2/3} -1\]

7 years ago
OpenStudy (anonymous):

does that make it clear?

7 years ago
OpenStudy (anonymous):

do you mean\[g(s)=s ^{\frac{1}{3}} - \frac{1}{s ^{\frac{2}{3}}-1}\]

7 years ago
OpenStudy (anonymous):

no, the whole numerator divided by the denominator

7 years ago
OpenStudy (anonymous):

oh so it is\[\frac {s ^{\frac{1}{3}}-1}{s ^{\frac{2}{3}}-1}\]

7 years ago
OpenStudy (anonymous):

correct?

7 years ago
OpenStudy (anonymous):

in that case you should use the quotient rule

7 years ago
OpenStudy (anonymous):

yessss! so can you tell me the answer..

7 years ago
OpenStudy (anonymous):

i got \[\frac{-1-s ^{\frac{2}{3}}+2s ^{\frac{1}{3}}}{3s ^{\frac{2}{3}} (s ^{\frac{2}{3}}-1)^{2}}\]

7 years ago
OpenStudy (anonymous):

how did you get it? explain please..

7 years ago
OpenStudy (anonymous):

I just used the quotient rule and simplified.

7 years ago
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