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

7 years ago(1/3) s^-2/3 + (2/3) s^(-5/3)

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

7 years ago1/s^2/3 = s^(-2/3)

7 years ago-2/3-1 = -5/3

7 years agooh okay. but what about the rule? it should be quotient.

7 years agoit doesnt matter. it is easier my way

7 years agoI 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 agowait. the question is ambiguous. can you post your question using the equation button below?

7 years agotype in frac{1}{s} to get 1/s

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

7 years agodoes that make it clear?

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

7 years agono, the whole numerator divided by the denominator

7 years agooh so it is\[\frac {s ^{\frac{1}{3}}-1}{s ^{\frac{2}{3}}-1}\]

7 years agocorrect?

7 years agoin that case you should use the quotient rule

7 years agoyessss! so can you tell me the answer..

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

7 years agohow did you get it? explain please..

7 years agoI just used the quotient rule and simplified.

7 years ago