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Find $ \frac{dy}{dx}\;when\;y=\Large x^{x^{x^{x.....}}}$

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Toolbox:
  • Express $ y = x^y$
  • Take log before differentiating.
$ y = x^y$
$ log\: y = y\: log\: x$
 
$ \large\frac{1}{y}\: \large\frac{dy}{dx}=\large\frac{y}{x}+ log\: x \large\frac{dy}{dx}$
$ \large\frac{dy}{dx}=\large\frac{y^2}{x(1-y\: log\: x)}$

 

answered Mar 12, 2013 by thanvigandhi_1
edited Mar 26, 2013 by thanvigandhi_1
 
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