Please help me calculate the first derivate of the function $f(x)=x^{x^{x}}$. Thanks.
How to find the derivate of the function $f(x)=x^{x^{x}}$
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1Is $f(x)=x^{(x^x)}$ or $f(x)=(x^x)^x$? – 2012-09-26
2 Answers
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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0It 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. – 2012-09-26
Hint:
Try calculating $\log{f(x)}$