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Question-137282




Question Number 137282 by mathlove last updated on 31/Mar/21
Answered by EDWIN88 last updated on 31/Mar/21
(d/dx)(fofofof)(x)=f ′(x)f ′(f(x))f ′(f(f(x))) f ′(f(f(f(x))))
$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{fofofof}\right)\left(\mathrm{x}\right)=\mathrm{f}\:'\left(\mathrm{x}\right)\mathrm{f}\:'\left(\mathrm{f}\left(\mathrm{x}\right)\right)\mathrm{f}\:'\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right)\:\mathrm{f}\:'\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right)\right) \\ $$
Answered by floor(10²Eta[1]) last updated on 31/Mar/21
(fofofof)(x)=g(x)=e^e^e^e^x      g′(x)=e^e^e^e^x    e^e^e^x   e^e^x  e^x
$$\left(\mathrm{fofofof}\right)\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{e}^{\mathrm{e}^{\mathrm{e}^{\mathrm{x}} } } } \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{e}^{\mathrm{e}^{\mathrm{e}^{\mathrm{x}} } } } \mathrm{e}^{\mathrm{e}^{\mathrm{e}^{\mathrm{x}} } } \mathrm{e}^{\mathrm{e}^{\mathrm{x}} } \mathrm{e}^{\mathrm{x}} \\ $$

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