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Math Help - Differentiating the real log

  1. #1
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    Differentiating the real log

    Find the Derivative:
    h(x) = \frac{\log_{10}(x)}{x}

    the answer is \frac{1}{x^2}(\log_{10}\frac{e}{x})
    How??
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the Derivative:
    h(x) = \frac{\log_{10}(x)}{x}

    the answer is \frac{1}{x^2}(\log_{10}\frac{e}{x})
    How??
    You need the change of base formula for logarithms:

    \log_a(b)=\frac{\log_c(b)}{\log_c(a)},

    or in this case:

    \log_{10}(x)=\frac{\log_e(x)}{\log_e(10)}.

    So:

    h(x) = \frac{\log_{e}(x)}{x\,\log_e(10)}.

    This may be differentiated using the quotiont rule (or the product rule):

    \frac{dh}{dx}=\frac{1}{\log_e(10)}\left( -\frac{\log_e(x)}{x^2}+\frac{1}{x^2}\right)

    ...... =\frac{1}{\log_e(10)}\left( -\frac{\log_e(x)}{x^2}+\frac{\log_e(e)}{x^2}\right)  =\frac{1}{x^2}\left(\frac{ \log_{e}(e/x)}{\log_e(10)} \right)==\frac{1}{x^2}\left({ \log_{10}(e/x)} \right)

    RonL
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the Derivative:
    h(x) = \frac{\log_{10}(x)}{x}

    the answer is \frac{1}{x^2}(\log_{10}\frac{e}{x})
    How??
    h(x) = Log(x) /x ------where "Log" is to the base 10.

    Notes:
    -----d/dx Log(base b)[u] = [1 /ln(b)]*(1/u) du/dx

    -----To change log(base b)[A] to log(base c), then:
    log(base b)[A] = log(base c)[A] / log(base c)[b]

    So,
    h'(x) = {x[(1/ln(10) *1/x] -Log(x)} /x^2
    h'(x) = {1/ln(10) -Log(x)} /x^2

    Convert the ln(10) to Log(10),

    h'(x) = {[1 / Log(10)/Log(e)] -Log(x)} /x^2

    h'(x) = {(Log(e) /Log(10)) -Log(x)} /x^2
    h'(x) = {(Log(e) /1) -Log(x)} /x^2
    h'(x) = {Log(e) -Log(x)} /x^2
    h'(x) = Log(e/x) /x^2 ---------------answer.
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