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calculus

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how to find the second derivative of quotient rule?
f(x): 700v^2 + 3450/v
f'(x) : 700v^2 -3450/v^2
f"'(x) : ???

  • calculus - ,

    sorry typo error above, question is,

    f(x): 700v^2 + 34500/v
    f'(x) : 700v^2 -34500/v^2
    f"'(x) : ???

  • calculus - ,

    the way you typed it ....
    f'(x) = 1400v - 3450/v^2 or 1400v - 3450v^-2

    f''(x) =1400 + 6900v^-3

    IF you meant f(x) = (700v^2 + 3450)/v
    I would change it to
    f(x) = 700v + 3400v^-1
    f'(x) = 700 - 3400v^-2
    f''(x) = 6800v^-3 or 6800/v^2

  • calculus - ,

    The question should be:


    f(x): 700v^2 + 34500/v
    f'(x) : 700v^2 -34500/v^2
    f''(x) : ???

  • calculus - ,

    You missed my point, the issue was not whether it was 3450 or 34500, the point was the use of brackets.

    Did you not look at my reply?
    simply change 3450 to 34500 and follow the steps.

    You derivative would be correct if you had placed it in brackets , such as

    f'(x) = (700v^2 - 34500)/v^2

    which would reduce to
    f'(x) = 700 - 34500/v^2 = 700 - 34500v^-2

    then f''(x) = 69000/v^3

  • calculus - ,

    pls ignore the above question...

    to make it clearer for the question, it is...

    f(x): (700v^2 + 34500)/v
    f'(x) : (700v^2 -34500)/v^2
    f''(x) : ???

  • calculus - ,

    i got the answer!! thank you!!!

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