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y(s)=1/5s^2 - 6/s find y'(s)
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is s^2 in the denom? and s too?
let me rewrite
is this your equation:\[y(s)=\frac{1}{5s^2}-\frac{6}{s}\]
y(s)=(1)/(5s^2) - (6)/(s)
YES!
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sorry about the confusion everyone
i see...well the constants don't affect the differentiation..just turn the s to negative exponents then power rule. try
you also dont have to LCD it or anything...itll just make it harder
yes - Igbasallote is correct. rewrite the equation as:\[y(s)=\frac{1}{5}s^{-2}-6s^{-1}\]you should then be able to use the standard rules of differentiation.
i.e. use the rule that if:\[y=ax^b\]then:\[y'=abx^{b-1}\]
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do you understand?
YES IT DOES! THANKS!
yw
:D
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