Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

-2/3-1 = -5/3

Answer 5

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

Answer 6

it doesnt matter. it is easier my way

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

does that make it clear?

Answer 12

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

Answer 13

no, the whole numerator divided by the denominator

Answer 14

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

Answer 15

correct?

Answer 16

in that case you should use the quotient rule

Answer 17

yessss! so can you tell me the answer..

Answer 18

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

Answer 19

how did you get it? explain please..

Answer 20

I just used the quotient rule and simplified.

Answer 21

http://www.math.hmc.edu/calculus/tutorials/quotient_rule/