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

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

Mathematics
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\[the derivative = 2x ^{-3}-5x ^{-1}\]
an then you do this to derive it....
\[derivative= -6x+5x ^{-2}\]

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the above is your answer
or you can write the answer like this...
\[derivative = -6x+(5/x ^{2})\]
Fan me if that helped :D
why does the 5/s change to 5s^-1?? can you please tell me how to change
There 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
other examples are 288/x^7 = 288x^-7
IF 5/s if equal to 5s^-1 then why is it 5s^-2??
actually, its the rules of exponents. So when you have \[X ^{-a} = 1/(X ^{a})\]
the negative sign represents the inverse.
What 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 -> (-3-1)= -4
wow I just realised that I made a mistake
the final answer should be -6x^(-4)+5x^(-2)
sorry about that
yeah so like I was saying before you subtract 1 from the exponents
the final answer in the ANSWERS SHEET IS \[-6/s^{4} + 5/s^{2}\]
in replying to your question above, \[r=(2/s ^{3}) - (5/s)\] \[r= 2s ^{-3} - 5s ^{-1}\] Find the derivative \[r'= (-3)(2s ^{-3-1}) - (-1)(5s ^{-1-1})\] \[r' = -6s ^{-4} + 5s ^{-2}\]
thank you heaps... that is great ~ i can actually see what is happening :)
no problem. calculus is very easy if you just follow all the rules properly =)
do u no a good website where i can learn calculus quickly?
Calculus 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.
great thanks heaps :)
yup

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