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Math Help - First Derivative

  1. #1
    Senior Member sakonpure6's Avatar
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    First Derivative

    we have: y= 10x ^{Ln10x}
    First thing we do: $ Ln (y) = Ln(10x) * Ln(10x) $
    Finally after applying the "rules" : $ y' = y* [ \frac {2 Ln (10x)}{x} ] $

    Am I correct? Because wolfram alpha gives the answer in a different form first derivative of y= 10x^(Ln10x) - Wolfram|Alpha
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  2. #2
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    Re: First Derivative

    Quote Originally Posted by sakonpure6 View Post
    we have: y= 10x ^{Ln10x}
    First thing we do: $ Ln (y) = Ln(10x) * Ln(10x) $
    Finally after applying the "rules" : $ y' = y* [ \frac {2 Ln (10x)}{x} ] $

    Am I correct? Because wolfram alpha gives the answer in a different form first derivative of y= 10x^(Ln10x) - Wolfram|Alpha

    $y=10 x^{\ln(10x)}=10 \exp(\ln(x)\ln(10x))$

    $\dfrac {dy}{dx}=10\exp(\ln(x)\ln(10x))\times \dfrac {d}{dx}\left(\ln(x)\ln(10x)\right)=$

    $10\exp(\ln(x)\ln(10x))\times\left(\dfrac 1 x \ln(10x)+\ln(x)\dfrac{10}{x}\right)=$

    $10^{\ln(10x)} \left(\dfrac 1 x \ln(10x)+\ln(x)\dfrac{10}{x}\right)$

    This is what Mathematica gives.. I 'm sure this an the various forms wolfram alpha gives are all equivalent.
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  3. #3
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    Re: First Derivative

    I presume that "log10x" means "log, base 10, of x" or log_{10}(x). However, taking the logarithm of both sides does NOT give log_{10}(x)\log_{10}(x). it gives, rather, log_{10}(y)= log_{10}(10)+ log_{10}(x^{log_{10}(10x)= 1+ log_{10}(x)log_{10}(10x}= 1+ log_{10}(x)+ log_{10}(x)log_{10}(x)= 1+ log_{10}(x)+ (log_{10}(x))^2.

    The derivative of both sides of that, with respect to x, is \frac{1}{y}\frac{dy}{dx}= \frac{ln(10)}{x}+ \frac{2ln(10)(log_{10}(x))}{x}
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