Here's the question you clicked on:
Dave321
Given: v=(pi)(12r^2-3r^3) v'(r)=(pi)(24r-9r^2) My question is; why doesn't the derivative of pi=0, since it's a constant?
v=(pi)(12r^2-3r^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)
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
Or they could be going by product rule, so since v = a*b then v' = a*b' + a'*b' In this case, v' = pi*(24r-9r^2) + 0*(12r^2 - 3r^3)
Thanks people. Not sure why I didn't see that...