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Math Help - 2 derivative problems

  1. #1
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    2 derivative problems

    Find the derivative of ln e^u/u
    the answer is 0 but I don't know how to get it





    lim
    h-> 0 square root of (25-h) -5/h


    The answer is 1/10 but I once again don't know how to do it.
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  2. #2
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    If it's supposed to be 0 then you must mean y= \frac{\ln(e^u)}{u}

    y=\frac{u\ln(e)}{u} by log rules

    y=\frac{u*1}{u}=1

    y'=0
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  3. #3
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    Quote Originally Posted by Jameson View Post
    If it's supposed to be 0 then you must mean y= \frac{\ln(e^u)}{u}

    y=\frac{u\ln(e)}{u} by log rules

    y=\frac{u*1}{u}=1

    y'=0

    Thank you for the help, can you help me with the second problem?
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  4. #4
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    If I'm reading it right it's \lim_{h \rightarrow 0} \frac{\sqrt{25-h}-5}{h}

    Multiply the numerator and denominator by \sqrt{25-h}+5

    \frac{\sqrt{25-h}-5}{h} * \frac{\sqrt{25-h}+5}{\sqrt{25-h}+5}

    This comes out to be \frac{-h}{h(\sqrt{25-h}+5)}=\frac{-1}{\sqrt{25-h}+5}

    Now try the limit again.
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  5. #5
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    Quote Originally Posted by Jameson View Post
    If I'm reading it right it's \lim_{h \rightarrow 0} \frac{\sqrt{25-h}-5}{h}

    Multiply the numerator and denominator by \sqrt{25-h}+5

    \frac{\sqrt{25-h}-5}{h} * \frac{\sqrt{25-h}+5}{\sqrt{25-h}+5}

    This comes out to be \frac{-h}{h(\sqrt{25-h}+5)}=\frac{-1}{\sqrt{25-h}+5}

    Now try the limit again.
    it came out -1/10, but the answer is 1/10

    did I do something wrong there?
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  6. #6
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    No, I'm pretty sure the answer is negative. I just checked the limit on a calculator and it confirms it's -1/10.
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