Question Number 206882 by efronzo1 last updated on 29/Apr/24
$$\:\:{f}\left({x}\right)=\mathrm{tan}\:^{\mathrm{2}} {x}\:\sqrt{\mathrm{tan}\:{x}\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}\sqrt[{\mathrm{4}}]{\mathrm{tan}\:{x}\sqrt[{\mathrm{5}}]{\mathrm{tan}\:{x}\sqrt{…}}}}} \\ $$$$\:{f}\:'\left(\frac{\pi}{\mathrm{4}}\right)=? \\ $$
Answered by MM42 last updated on 29/Apr/24
$${f}\left({x}\right)={tan}^{\mathrm{2}} {x}×{tan}^{\frac{\mathrm{1}}{\mathrm{2}}} {x}×{tan}^{\frac{\mathrm{1}}{\mathrm{6}}} {x}×{tan}^{\frac{\mathrm{1}}{\mathrm{24}}} ×{tan}^{\frac{\mathrm{1}}{\mathrm{120}}} ×… \\ $$$$=\left({tanx}\right)_{} ^{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{24}}+…+\frac{\mathrm{1}}{{n}!}+…\:\:\:} =\left({tanx}\right)^{{e}} \\ $$$$\Rightarrow{f}^{\:'} \left({x}\right)={e}\left({tanx}\right)^{{e}−\mathrm{1}} ×\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right) \\ $$$$\Rightarrow{f}^{\:'} \left(\frac{\pi}{\mathrm{4}}\right)=\mathrm{2}{e}\:\:\checkmark \\ $$$$ \\ $$