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Math Help - derivative check

  1. #1
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    derivative check

    y = \frac{e^xlogx}{log(2x+1)}

     <br />
y '= \frac{e^x(logx+\frac{1}{x})log(2x+1)-(\frac{2}{2x+1})(e^xlogx)}{(log(2x+1))^2}<br />

    can i then cancel out the log(2x+1) from the top and one from the bottom leaving the denominator as log(2x+1) ?? thank u
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  2. #2
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    This is a much easier way...I don't think you took your derivative right because the derivative of logx is not 1/x. It would be right if you change your logs to ln.
    y = \frac{e^xlogx}{log(2x+1)}

    y = \frac{\frac{e^xlnx}{ln10}}{\frac{ln(2x+1)}{ln10}}

    y = \frac{e^xlnx}{ln(2x+1)}

    y '= \frac{e^x(lnx+\frac{1}{x})ln(2x+1)-(\frac{2}{2x+1})(e^xlnx)}{(ln(2x+1))^2}<br />

    I don't see anything you can cancel out.
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  3. #3
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    Quote Originally Posted by Linnus View Post
    This is a much easier way...I don't think you took your derivative right because the derivative of logx is not 1/x. It would be right if you change your logs to ln.
    y = \frac{e^xlogx}{log(2x+1)}

    y = \frac{\frac{e^xlnx}{ln10}}{\frac{ln(2x+1)}{ln10}}

    y = \frac{e^xlnx}{ln(2x+1)}

    y '= \frac{e^x(lnx+\frac{1}{x})ln(2x+1)-(\frac{2}{2x+1})(e^xlnx)}{(ln(2x+1))^2} <br />

    I don't see anything you can cancel out.
    so the ln(2x+1) dont cancel out ?

    on this website (solving derivatives step-by-step) says the differentiation of logx is 1/x ? even on my answer paper for this question it says 1/x
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  4. #4
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    Quote Originally Posted by jvignacio View Post
    so the ln(2x+1) dont cancel out ?

    on this website (solving derivatives step-by-step) says the differentiation of logx is 1/x ? even on my answer paper for this question it says 1/x
     \frac {dy}{dx} lnx=\frac {1}{x}
     \frac {dy}{dx} log_bx=\frac {1}{x}lnb
    No it doesn't cancel out because you cannot factor out a ln(2x+1) on the top.

    It says on the website "On this site log[x] is the same as ln[x]"
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  5. #5
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    Quote Originally Posted by jvignacio View Post
    so the ln(2x+1) dont cancel out ?

    on this website (solving derivatives step-by-step) says the differentiation of logx is 1/x ? even on my answer paper for this question it says 1/x
    If \log represents \log_e then the derivative of \log x is \frac{1}{x}.

    Note: \log_e is usually represented by \ln . Hence the comments previously posted.

    To answer your question, consider ..... Can you take \log (2x + 1) out as a common factor in the numerator? No? Then you can't cancel it out.
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  6. #6
    gml
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    \frac {dy}{dx} log_bx=\frac {1}{x}lnb
    This seems a little bit misleading but maybe I am reading it poorly.
    dy/dx(logx) is actually 1/x * 1/lnx, not 1/x * lnx which would put the lnx in the numerator instead of the denominator.
    Last edited by mr fantastic; January 10th 2009 at 02:07 PM. Reason: Fixed the latex
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  7. #7
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    Quote Originally Posted by gml View Post
    \frac {dy}{dx} log_bx=\frac {1}{x}lnb
    This seems a little bit misleading but maybe I am reading it poorly.
    dy/dx(logx) is actually 1/x * 1/lnx, not 1/x * lnx which would put the lnx in the numerator instead of the denominator.
    Did you mean \frac{d \log x}{dx}= (1/x) (1/\ln(10))? It is certainly NOT (1/x) (1/\ln(x)).

    If y= \log(x) then x= 10^y= e^{\ln(10^y)}= e^{yln(10)} so, by the chain rule, \frac{dx}{dy}= \ln(10) e^{y \ln (10)}= \ln(10) 10^y= \ln (10) x. Thus, if y= \log (x), \frac{dy}{dx}= \frac{1}{x \ln(10)}.
    Last edited by mr fantastic; January 10th 2009 at 07:44 PM. Reason: Fixed some latex
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