Question

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?

Answer 1

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)

Answer 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

Answer 3

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)

Answer 4

yep

Answer 5

Thanks people. Not sure why I didn't see that...