# How to find the derivate of the function $f(x)=x^{x^{x}}$

Please help me calculate the first derivate of the function $f(x)=x^{x^{x}}$. Thanks.

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Hint: Rewrite your function as $f(x) = \exp\log x^{x^x} = \exp(x^x \log x) = \exp(\exp(\log x^x)\log x) = \exp(\exp(x\log x)\log x)$ – martini Sep 26 '12 at 19:27
Is $f(x)=x^{(x^x)}$ or $f(x)=(x^x)^x$? – poirot Sep 26 '12 at 21:37

Implement the formula:

1) $\log_a x^n=n\log_a x$

2) $\ln f(x)=\frac{1}{f(x)}\cdot f'(x)$

3) $[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)$

4) $[\ln x]'=\frac{1}{x}$

5) $a^{x_1}:a^{x_2}=\frac{a^{x_{1}}}{a^{x_{2}}}=a^{x_1-x_2}$

6) $x^{-n}=\frac{1}{x^n}$

Now turn find the derivate of this function.

$f(x)=x^{x^{x}}$/$\cdot$$\ln \ln f(x)=\ln x^{x^{x}} \ln f(x)=x^x\ln x / ^' [\ln f(x)]'=(x^x\ln x)' \frac{1}{f(x)}\cdot f'(x)=(x^x\ln x)' \frac{1}{f(x)}\cdot f'(x)=(x^x)'\ln x + x^x(\ln x)' \frac{1}{f(x)}\cdot f'(x)=(x^x)'\ln x + x^x\cdot\frac{1}{x} \frac{1}{f(x)}\cdot f'(x)=(x^x)'\ln x + x^x\cdot x^{-1} \frac{1}{f(x)}\cdot f'(x)=(x^x)'\ln x + x^{x-1} / \cdot f(x) f(x)\cdot\frac{1}{f(x)}\cdot f'(x)=f(x)[(x^x)'\ln x + x^{x-1}] f'(x)=f(x)[(x^x)'\ln x + x^{x-1}] f'(x)=f(x)[f_1(x)\ln x + x^{x-1}] Now find the derivate of the function f_1(x)=x^x. f_1(x)=x^x/\cdot$$\ln$

$\ln f_1(x)=\ln x^x$

$\ln f_1(x)=x\ln x$/'

$[\ln f_1(x)]'=[x\ln x]'$

$\frac{1}{f_1(x)}\cdot f'_1(x)=x'\ln x+x(\ln x)'$

$\frac{1}{f_1(x)}\cdot f'_1(x)=\ln x+x\cdot\frac{1}{x}$

$\frac{1}{f_1(x)}\cdot f'_1(x)=\ln x+1$/$\cdot$ $f_1(x)$

$f'_1(x)=f_1(x)(\ln x+1)$

$f'_1(x)=(x^x)'=x^x(\ln x+1)$

Definitly:

$f'(x)=(x^{x^{x}})'=f(x)[f_1(x)\ln x + x^{x-1}]$

$f'(x)=(x^{x^{x}})'=x^{x^{x}}[x^x(\ln x+1)\ln x + x^{x-1}]$

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It is "D-E-F-I-N-I-T-E-L-Y". It means, losely, "for sure". You might want to be using "Finally" or "All things considered" instead. – Pedro Tamaroff Sep 26 '12 at 20:37

# Hint:

Try calculating $\log{f(x)}$

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