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Question Number 201534 by Mathspace last updated on 08/Dec/23
let f(x)=tanx  find f^((n)) (x) with n integr  natural
$${let}\:{f}\left({x}\right)={tanx} \\ $$$${find}\:{f}^{\left({n}\right)} \left({x}\right)\:{with}\:{n}\:{integr} \\ $$$${natural} \\ $$
Commented by Frix last updated on 09/Dec/23
There′s a very complicated formula, you must  search the www  But we can give a nice sequence  t=tan x  u=at^b  ⇒ u′=(t^2 +1)(df/dt)=ab(t^(b+1) +t^(b−1) )  If we start with u_0 =t (a=b=1)  u_0 =t  u_1 =t^2 +1  u_2 =(t^2 +1)×2t=2t^3 +2t  u_3 =(t^2 +1)(6t^2 +2)=6t^4 +8t^2 +2  u_3 =(t^2 +1)(24t^3 +16t)=24t^5 +40t^3 +16t  u_4 =(t^2 +1)(120t^4 +120t^2 +16)=       =120t^6 +240t^4 +136t^2 +16  ...
$$\mathrm{There}'\mathrm{s}\:\mathrm{a}\:\mathrm{very}\:\mathrm{complicated}\:\mathrm{formula},\:\mathrm{you}\:\mathrm{must} \\ $$$$\mathrm{search}\:\mathrm{the}\:\mathrm{www} \\ $$$$\mathrm{But}\:\mathrm{we}\:\mathrm{can}\:\mathrm{give}\:\mathrm{a}\:\mathrm{nice}\:\mathrm{sequence} \\ $$$${t}=\mathrm{tan}\:{x} \\ $$$${u}={at}^{{b}} \:\Rightarrow\:{u}'=\left({t}^{\mathrm{2}} +\mathrm{1}\right)\frac{{df}}{{dt}}={ab}\left({t}^{{b}+\mathrm{1}} +{t}^{{b}−\mathrm{1}} \right) \\ $$$$\mathrm{If}\:\mathrm{we}\:\mathrm{start}\:\mathrm{with}\:{u}_{\mathrm{0}} ={t}\:\left({a}={b}=\mathrm{1}\right) \\ $$$${u}_{\mathrm{0}} ={t} \\ $$$${u}_{\mathrm{1}} ={t}^{\mathrm{2}} +\mathrm{1} \\ $$$${u}_{\mathrm{2}} =\left({t}^{\mathrm{2}} +\mathrm{1}\right)×\mathrm{2}{t}=\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t} \\ $$$${u}_{\mathrm{3}} =\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{6}{t}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{6}{t}^{\mathrm{4}} +\mathrm{8}{t}^{\mathrm{2}} +\mathrm{2} \\ $$$${u}_{\mathrm{3}} =\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{24}{t}^{\mathrm{3}} +\mathrm{16}{t}\right)=\mathrm{24}{t}^{\mathrm{5}} +\mathrm{40}{t}^{\mathrm{3}} +\mathrm{16}{t} \\ $$$${u}_{\mathrm{4}} =\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{120}{t}^{\mathrm{4}} +\mathrm{120}{t}^{\mathrm{2}} +\mathrm{16}\right)= \\ $$$$\:\:\:\:\:=\mathrm{120}{t}^{\mathrm{6}} +\mathrm{240}{t}^{\mathrm{4}} +\mathrm{136}{t}^{\mathrm{2}} +\mathrm{16} \\ $$$$… \\ $$

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