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Math Help - Differentiation of Log functions

  1. #1
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    Differentiation of Log functions

    Sigh...calculus is giving me a hard time...Trig was much easier. Anyone recommend a good calculus book to help me put things in perspective? Anyways, I need help with differentiating the following log functions:

    1. f(x)=\sqrt{x} lnx
    2. f(u)=\frac{lnu}{1+ln(2u)}

    thanks
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  2. #2
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    Quote Originally Posted by c_323_h
    1. f(x)=\sqrt{x} lnx
    First the product rule,
    (\sqrt{x})'\ln x+\sqrt{x}(\ln x)'
    Thus,
    \frac{\ln x}{2\sqrt{x}}+\frac{\sqrt{x}}{x}
    Thus,
    \frac{\ln x}{2\sqrt{x}}+\frac{1}{\sqrt{x}}
    Thus,
    \frac{1}{\sqrt{x}} (1/2\ln x+1)
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  3. #3
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    Quote Originally Posted by c_323_h
    Anyone recommend a good calculus book to help me put things in perspective?
    The publishers that I love,
    Prentice Hall and McGraw-Hill search their sites.
    (The books are very expensive).

    So you might be tempted to buy,
    Calculus for Dummies or Schaum's Outline- I think these publishers are horrific and stay away from them.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker
    The publishers that I love,
    Prentice Hall and McGraw-Hill search their sites.
    (The books are very expensive).
    textboooks! you're freakin psycho! just kidding...i was thinking more along the lines of "A Tour of the Calculus". i used the product rule but my answer wasn't the same as the answer in my book..hmm maybe the book is wrong or i didn't simplify it correctly.
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  5. #5
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    Hello, c_323_h!

    I tried something on #2
    . . and found it can be simplified beyond all recognition . . .


    2)\;\;f(u)\;=\;\frac{\ln u}{1+\ln2u}

    Since \ln(2u) \:=\:\ln(2) + \ln(u), we have:

    . . f(u)\;=\;\frac{\ln u }{1 + \ln2 + \ln u}


    Differentiate:

    . . f'(u) \;= \;\frac{[1 + \ln 2 + \ln u]\cdot\frac{1}{u} - \ln u\cdot\frac{1}{u}}{(1 + \ln2 + \ln u)^2}


    Multiply top and bottom by u:

    . . f'(u) \;= \;\frac{1 + \ln 2 + \ln u - \ln u}{u(1 + \ln 2 + \ln u)^2}


    Therefore: . f'(u) \;=\;\frac{1 + \ln 2}{u(1 + \ln2u)^2}


    If that's what they did, no wonder your answer is different!

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Let's see what they can do to #1 . . .

    1)\;\;f(x)\;=\;x^{\frac{1}{2}}\ln x

    Then: . f'(x)\;=\;\left(\frac{1}{2}x^{-\frac{1}{2}}\right)\!\cdot\!\ln x \,+ \,x^{\frac{1}{2}}\!\cdot\!\left(\frac{1}{x}\right) \;= \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}<br />

    Get a common denominator: . f'(x)\;=\;\frac{\ln x}{2\sqrt{x}} + \frac{2}{2\sqrt{x}} \;= \;\frac{\ln x + 2}{2\sqrt{x}}

    Last edited by Soroban; July 1st 2006 at 10:29 PM.
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  6. #6
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    Quote Originally Posted by Soroban
    Hello, c_323_h!

    I tried something on #2
    . . and found it can be simplified beyond all recognition . . .



    Since \ln(2u) \:=\:\ln(2) + \ln(u), we have:

    . . f(u)\;=\;\frac{\ln u }{1 + \ln2 + \ln u}


    Differentiate:

    . . f'(u) \;= \;\frac{[1 + \ln 2 + \ln u]\cdot\frac{1}{u} - \ln u\cdot\frac{1}{u}}{(1 + \ln2 + \ln u)^2}


    Multiply top and bottom by u:

    . . f'(u) \;= \;\frac{1 + \ln 2 + \ln u - \ln u}{u(1 + \ln 2 + \ln u)^2}


    Therefore: . f'(u) \;=\;\frac{1 + \ln 2}{u(1 + \ln2u)^2}


    If that's what they did, no wonder your answer is different!

    THANK YOU! That's the exact answer as in my book. So that's how they got it! Does anyone know of any other ways to solve this one?
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  7. #7
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    Hello again, c_323_h!

    Does anyone know of any other ways to solve this one?

    Since 1 + \ln(2u) \;=\;\ln(e) + \ln(2u) \;=\;\ln(2eu)

    . . we have: . f(u)\:=\:\frac{\ln(u)}{\ln(2eu)}


    Quotient Rule: . f'(u)\;=\;\frac{\ln(2eu)\cdot\frac{1}{u} - \ln(u)\cdot\frac{1}{2eu}\cdot2e}{[\ln(2eu)]^2}

    We have: . f'(u)\;=\;\frac{\frac{\ln(2eu)}{u} - \frac{\ln(u)}{u}}{\ln(2eu)]^2}

    Multiply top and bottom by u:\;\;f'(u) \;=\;\frac{\ln(2eu) - \ln(u)}{u[\ln(2eu)]^2}\;= \frac{\ln\left(\frac{2eu}{u}\right)}{u[\ln(2eu)]^2}

    Therefore: . f'(u) \;= \;\frac{\ln(2e)}{u[\ln(2eu)]^2]}

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  8. #8
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    thanks. what does differentiate with respect to x mean?
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  9. #9
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    Quote Originally Posted by c_323_h
    thanks. what does differentiate with respect to x mean?
    Nothing, just ignore that.
    Basically, just differentiate the function.
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