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Given: v=(pi)(12r^23r^3)
v'(r)=(pi)(24r9r^2)
My question is; why doesn't the derivative of pi=0, since it's a constant?
 2 years ago
 2 years ago
Given: v=(pi)(12r^23r^3) v'(r)=(pi)(24r9r^2) My question is; why doesn't the derivative of pi=0, since it's a constant?
 2 years ago
 2 years ago

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Wink_PinkBest ResponseYou've already chosen the best response.0
v=(pi)(12r^23r^3) = pi 12 r^2  pi 3 r^3 now V ' = pi 12( 2r)  pi 3 ( 3 r^2) = pi 24r pi 9 r^2 = pi( 24r 9 r^2)
 2 years ago

Wink_PinkBest ResponseYou've already chosen the best response.0
here pi is not a term all by itself if u had v = 12r^2 3r^3 +PI, then u treat the PI as wht u r thinking and drivative is o
 2 years ago

cgonzalez31Best ResponseYou've already chosen the best response.1
Or they could be going by product rule, so since v = a*b then v' = a*b' + a'*b' In this case, v' = pi*(24r9r^2) + 0*(12r^2  3r^3)
 2 years ago

Dave321Best ResponseYou've already chosen the best response.0
Thanks people. Not sure why I didn't see that...
 2 years ago
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