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Thread: Derivatives problem

  1. #1
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    Derivatives problem

    Hello, back again with another derivatives question:

    A grocery store determines that after t hours on the job, a new cashier can ring up N(t) = 20 - (30

    √(9 + t^2)

    items per minute. Find N' (t), the rate at which the cashier's productivity is changing.

    I've solved the derivative to

    (9 + t^2)^(1/2) - 30 t

    (9+t^2)^(3/2)

    But the correct answer should be simply

    30 t

    (9 + t^2)^(3/2)

    I have no idea how the (9 + t^2)^(1/2) in the numerator vanishes, though I've tried re-doing it several times. I'd greatly appreciate it if anyone could show me the steps to simplifying this derivative. Thank you in advance!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by starswept View Post
    Hello, back again with another derivatives question:

    A grocery store determines that after t hours on the job, a new cashier can ring up N(t) = 20 - (30

    √(9 + t^2)

    items per minute. Find N' (t), the rate at which the cashier's productivity is changing.

    I've solved the derivative to

    (9 + t^2)^(1/2) - 30 t

    (9+t^2)^(3/2)

    But the correct answer should be simply

    30 t

    (9 + t^2)^(3/2)

    I have no idea how the (9 + t^2)^(1/2) in the numerator vanishes, though I've tried re-doing it several times. I'd greatly appreciate it if anyone could show me the steps to simplifying this derivative. Thank you in advance!
    Look at it this way:
    N(t) = 20 - \frac{30}{\sqrt{9 + t^2}}

    N(t) = 20 - 30 \cdot (9 + t^2)^{-1/2}

    Now just apply the chain and power rules:
    N^{\prime}(t) = -30 \cdot -\frac{1}{2}(9 + t^2)^{-3/2} \cdot 2t

    N^{\prime}(t) = 30t \cdot (9 + t^2)^{-3/2}

    N^{\prime}(t) = \frac{30t}{(9 + t^2)^{3/2}}

    -Dan
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