y=x^(x^x)
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Here's a hint $\displaystyle \frac{d}{dx}(x^x) = x^x\left(\ln x + \frac{1}{x}\right) $
$\displaystyle x^x\, x^{x^x - 1} + x^{x^x}\, \ln\!\left(x\right)\, \left(x\, x^{x - 1} + x^x\, \ln\!\left(x\right)\right) $
thanks bro. did u substitute right?
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