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Math Help - Derivative help

  1. #1
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    Derivative help

    Hi, I have two questions with derivatives that I'm struggling with:

    1) If f(x)= (xlnx)^2, find all points at which the graph of f(x) has a horizontal tangent line.

    I really have no idea where to start with this one- how do you find the derivative when the entire ln, not just the function, is squared?

    2) Formula for pH is: pH = -ln (H) divided by ln (10).

    Ingredients are being added to tomatoes so that the concentration of hydrogen ions in the mixture is given by H(t) = 30-5t-25 (e^(-1/5) - 1) moles per litre. Determine the rate of change of the pH value with respect to time after 10 s.

    I'd greatly appreciate any help. Thank you!
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  2. #2
    Senior Member tukeywilliams's Avatar
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    So  f(x) = (x \ln x)^2 . Then we want to set  f'(x) = 0  .

    So  f'(x) = 2x \ln x \left(1 + \ln x \right) or  0 =2x \ln x \left(1 + \ln x \right) . So  x = 0,1 .

     \text{pH} = (\frac{1}{\ln 10}-\ln(30-5t-25(e^{-1/5}-1))).

    Then  \frac{d\text{PH}}{dt} = \frac{1}{\ln 10}\left(-\frac{1}{30-5t}(5) \right) . So at  t = 10 ,  \frac{d\text{PH}}{dt} = \frac{1}{4 \ln 10} .
    Last edited by tukeywilliams; July 25th 2007 at 06:19 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tukeywilliams View Post
    So  f(x) = (x \ln x)^2 . Then we want to set  f'(x) = 0  .

    So  f'(x) = 2x \ln x \left(1 + \ln x \right) or  0 =2x \ln x \left(1 + \ln x \right) . So  x = 0,1 .
    There's a solution missing from the above solved equation:
    0 =2x \ln x \left(1 + \ln x \right)

    Thus
    2x = 0 ==> x = 0
    or
    ln(x) = 0 ==> x = 1
    or
    1 + ln(x) = 0 ==> x = \frac{1}{e}


    Quote Originally Posted by tukeywilliams View Post
     \text{pH} = (\frac{1}{\ln 10}-\ln(30-5t-25(e^{-1/5}-1))).

    Then  \frac{d\text{PH}}{dt} = \frac{1}{\ln 10}\left(-\frac{1}{30-5t}(5) \right) . So at  t = 10 ,  \frac{d\text{PH}}{dt} = \frac{1}{4 \ln 10} .
    There's a slight typo in your pH formula and I'm not sure what happened when you took the derivative:
    pH = \frac{-ln(30-5t-25 (e^{-1/5} - 1))}{ln(10)}

    \frac{d(pH)}{dt} = -\frac{\frac{1}{30 - 5t - 25(e^{-1/5} - 1)} \cdot (-5)}{ln(10)} = \frac{5}{ln(10) \cdot (30 - 5t - 25(e^{-1/5} - 1))}

    Now plug in t = 10.
    \frac{d(pH)}{dt}(t = 10) = \frac{5}{ln(10) \cdot (30 - 50 - 25(e^{-1/5} - 1))} = \frac{5}{ln(10) \cdot (-20 - 25(e^{-1/5} - 1))}

    = -\frac{1}{ln(10) \cdot (4 + 5(e^{-1/5} - 1))} \approx 0.140382

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    There's a solution missing from the above solved equation:
    0 =2x \ln x \left(1 + \ln x \right)
    Thus
    2x = 0 ==> x = 0
    or
    ln(x) = 0 ==> x = 1
    or
    1 + ln(x) = 0 ==> x = \frac{1}{e}
    However x=0 is not in the domain of the function. So x=1/e and x=1 are the answers.
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