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Math Help - stuck on second x^x

  1. #1
    Member
    Joined
    Sep 2009
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    124

    stuck on second x^x

    \begin{gathered}<br />
  f(x) = {x^{{x^x}}} \hfill \\<br />
  \ln y = {x^x}\ln x \hfill \\<br />
  \frac{1}<br />
{y}y' = \frac{1}<br />
{x}*\frac{1}<br />
{x} \hfill \\ <br />
\end{gathered}<br />

    but for the x^x, ln is needed but is
    \ln \ln y = \ln ({x^x}\ln x)<br />

    okay?
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  2. #2
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    Hello, genlovesmusic09!

    You are correct . . .


    Differentiate: . y \:=\: x^{x^x}

    Take logs: . \ln(y) \;=\;\ln\left(x^{x^x}\right) \quad\Rightarrow\quad \ln(y)\;=\;x^x\cdot\ln(x)

    Take logs: . \ln\left[\ln(y)\right] \;=\;\ln\left[x^x\cdot\ln(x)\right] \;=\; \ln\left[x^x\right] + \ln\left[\ln(x)\right]

    And we have: . \ln\left[\ln(y)\right] \;=\;x\cdot\ln(x) + \ln\left[\ln(x)\right]


    Differentiate: . \frac{1}{\ln(y)}\!\cdot\!\frac{1}{y}\!\cdot\! y' \;\;=\;\;x\!\cdot\!\frac{1}{x} \;+\; 1\!\cdot\!\ln(x) \;+\; \frac{1}{\ln(x)}\!\cdot\!\frac{1}{x}

    . . . . . . . . . . . . \frac{y'}{y\!\cdot\!\ln(y)} \;=\;1 + \ln(x) + \frac{1}{x\!\cdot\!\ln(x)}

    . . . . . . . . . . . . . . . y' \;=\;y \cdot \ln(y)\cdot\bigg[1 + \ln(x) + \frac{1}{x\!\cdot\!\ln(x)} \bigg]

    . . . . . . . . . . . . . . . y' \;=\;x^{x^x}\cdot\ln\left(x^{x^x}\right)\cdot\bigg[1 + \ln(x) + \frac{1}{x\!\cdot\!\ln(x)} \bigg]

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