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if f(x) = integral from 0 to x of (sqrt(t^3+1)), what is f'(2)?
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recall: \[F(x)=\int_{0}^{x}f(t)dt\] \[F'(x)=f(x)\]
so, what is sqrt(t^3+1), when t=2?
+-3 right?
sqrt(9) .... we tend to take the primary outcome, so 3 is fine ... but yes :)
we cant have a slope of 3 and -3 at the same time
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so how do you know which one it is for sure?
|dw:1428014008141:dw|
oh right forgot, so it has to be 3
the sqrt function is only valid for non negative numbers
thx
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