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Question Number 206882 by efronzo1 last updated on 29/Apr/24

f(x)=tan^2 x (√(tan x((tan x((tan x((tan x(√(...))))^(1/5) ))^(1/4) ))^(1/3) )) f ′((π/4))=?

$$\:\:{f}\left({x}\right)=\mathrm{tan}\:^{\mathrm{2}} {x}\:\sqrt{\mathrm{tan}\:{x}\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}\sqrt[{\mathrm{4}}]{\mathrm{tan}\:{x}\sqrt[{\mathrm{5}}]{\mathrm{tan}\:{x}\sqrt{...}}}}} \\ $$$$\:{f}\:'\left(\frac{\pi}{\mathrm{4}}\right)=? \\ $$

Answered by MM42 last updated on 29/Apr/24

f(x)=tan^2 x×tan^(1/2) x×tan^(1/6) x×tan^(1/(24)) ×tan^(1/(120)) ×... =(tanx)_ ^(2+(1/2)+(1/6)+(1/(24))+...+(1/(n!))+... ) =(tanx)^e ⇒f^( ′) (x)=e(tanx)^(e−1) ×(1+tan^2 x) ⇒f^( ′) ((π/4))=2e ✓

$${f}\left({x}\right)={tan}^{\mathrm{2}} {x}×{tan}^{\frac{\mathrm{1}}{\mathrm{2}}} {x}×{tan}^{\frac{\mathrm{1}}{\mathrm{6}}} {x}×{tan}^{\frac{\mathrm{1}}{\mathrm{24}}} ×{tan}^{\frac{\mathrm{1}}{\mathrm{120}}} ×... \\ $$$$=\left({tanx}\right)_{} ^{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{24}}+...+\frac{\mathrm{1}}{{n}!}+...\:\:\:} =\left({tanx}\right)^{{e}} \\ $$$$\Rightarrow{f}^{\:'} \left({x}\right)={e}\left({tanx}\right)^{{e}−\mathrm{1}} ×\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right) \\ $$$$\Rightarrow{f}^{\:'} \left(\frac{\pi}{\mathrm{4}}\right)=\mathrm{2}{e}\:\:\checkmark \\ $$$$ \\ $$

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