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Just want to check whether my answer is correct for quotient rule

F(v) : (600^3/2 + 20250)/v
F'(v) : (300^3/2 - 20250)/v^2
F"(v) : (-150v^5/2 + 40500)/v^3

Can anyone let me know whether my answer is correct for first and second derivative based on F(v) ??

  • calculus -

    Steve already went through this with you in

    Look at
    F(v) : (600^3/2 + 20250)/v

    what we are saying is that we suspect that that term would be 600v^(3/2)

    I got (-150v^(3/2) + 40500)/v^3

    As I told you in an earlier post, I would have simplified the original to
    f(x) = 600 v^(1/2) + 20250/v and I would not have used the quotient rule, but our answer should come out the same .

    My way: ....

    f(x) = 600 v^(1/2) + 20250 v^-1
    f'(x) = (1/2)600v^(-1/2) - 20250v^-2
    = 300 v^(-1/2) - 20250v^-2
    which is what you would get if we simplified your answer.

    then f''(x) = -150v^(-3/2) + 40500v^-3

    Your answer should have been
    (-150v^(3/2) + 40500)/v^3

  • calculus -

    Ok but the power 5/2 is wrong? Should have been power 3/2?

    I keep on getting the answer 5/2.. I don't know which part of my working is wrong

  • calculus -

    I do not want the simplified answer by the way...I need the answer as an expression only

  • calculus -

    based on my question, i retyped...
    F(v) : (600v^3/2 + 20250)/v
    F'(v) : (300v^3/2 - 20250)/v^2
    F"(v) : (-150v^5/2 + 40500)/v^3

    Reiny, isn't your answer suppose to be
    f(x) = 600 v^(3/2) + 20250 v^-1,
    instead of f(x) = 600 v^(1/2) + 20250 v^-1 ??

    1/2 power and 3/2 power there's a difference....

  • calculus -

    hi i think i know what you mean already.
    my answer for F"(v)
    (-150v^5/2 + 40500v)/v^4

    i simplified further, it became
    (-150v(v^3/2 - 270))/v^4
    cancel off the common factor v
    it became (-150(v^3/2 - 270))/v^3

    Am i correct??

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