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Question Number 137282 by mathlove last updated on 31/Mar/21

Answered by EDWIN88 last updated on 31/Mar/21

$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{fofofof}\right)\left(\mathrm{x}\right)=\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)\mathrm{f}{}^{\prime }\left(\mathrm{f}\left(\mathrm{x}\right)\right)\mathrm{f}{}^{\prime }\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right)\mathrm{f}{}^{\prime }\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right)\right)$$

Answered by floor(10²Eta[1]) last updated on 31/Mar/21

$$\left(\mathrm{fofofof}\right)\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)={\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{x}}}}}$$$${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)={\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{x}}}}}{\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{x}}}}{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{x}}}{\mathrm{e}}^{\mathrm{x}}$$

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