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Dave321

  • 4 years ago

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?

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  1. Wink_Pink
    • 4 years ago
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    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)

  2. Wink_Pink
    • 4 years ago
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    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

  3. cgonzalez31
    • 4 years ago
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    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)

  4. Wink_Pink
    • 4 years ago
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    yep

  5. Dave321
    • 4 years ago
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    Thanks people. Not sure why I didn't see that...

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