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anonymous
 5 years ago
Im getting confused with this::::
find the derivative of:
r = 2/s^3  5/s
anonymous
 5 years ago
Im getting confused with this:::: find the derivative of: r = 2/s^3  5/s

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[the derivative = 2x ^{3}5x ^{1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0an then you do this to derive it....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[derivative= 6x+5x ^{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the above is your answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or you can write the answer like this...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[derivative = 6x+(5/x ^{2})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Fan me if that helped :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why does the 5/s change to 5s^1?? can you please tell me how to change

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There is a rule with derivatives that says that if there is for example 5/s then you can write it as well as 5s^1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0other examples are 288/x^7 = 288x^7

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0IF 5/s if equal to 5s^1 then why is it 5s^2??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually, its the rules of exponents. So when you have \[X ^{a} = 1/(X ^{a})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the negative sign represents the inverse.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What I did there was following that rule where you multiply the exponent by the coefficient. so what I did was get =2x^(−3)−5x^(−1) using the rule of exponents. Then I multiplied the 2 in front of the x by the x's exponent (2X(3)) = 6 then I subtracted 1 from the exponent > (31)= 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow I just realised that I made a mistake

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the final answer should be 6x^(4)+5x^(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah so like I was saying before you subtract 1 from the exponents

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the final answer in the ANSWERS SHEET IS \[6/s^{4} + 5/s^{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in replying to your question above, \[r=(2/s ^{3})  (5/s)\] \[r= 2s ^{3}  5s ^{1}\] Find the derivative \[r'= (3)(2s ^{31})  (1)(5s ^{11})\] \[r' = 6s ^{4} + 5s ^{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you heaps... that is great ~ i can actually see what is happening :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no problem. calculus is very easy if you just follow all the rules properly =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do u no a good website where i can learn calculus quickly?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Calculus is actually something best learned in the setting of a classroom. But if you have any questions to ask just ask here and ill be more than happy to assist you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0great thanks heaps :)
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