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Question Number 124066 by Bird last updated on 30/Nov/20

let f(x)=arctan((2/x)) calculate f^((n)) (x) and f^((n)) (1) find f^((7)) ((1/7))

$${let}\:{f}\left({x}\right)={arctan}\left(\frac{\mathrm{2}}{{x}}\right) \\ $$$${calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{1}\right) \\ $$$${find}\:{f}^{\left(\mathrm{7}\right)} \left(\frac{\mathrm{1}}{\mathrm{7}}\right) \\ $$

Answered by Olaf last updated on 30/Nov/20

f(x) = arctan((2/x)) f′(x) = −(2/x^2 )×(1/(1+(4/x^2 ))) = −(2/(x^2 +4)) f′(x) = (i/2)[(1/(x−2i))−(1/(x+2i))] f^((n)) (x) = (i/2)(−1)^(n−1) (n−1)![(1/((x−2i)^n ))−(1/((x+2i)^n ))] f^((n)) (x) = (i/2)(−1)^(n−1) (n−1)![(1/(((√(x^2 +4))e^(−iarctan(2/x)) )^n ))−(1/(((√(x^2 +4))e^(iarctan(2/x)) )^n ))] f^((n)) (x) = ((i(−1)^(n−1) (n−1)!)/(2(x^2 +4)^(n/2) ))[(1/e^(−inarctan(2/x)) )−(1/e^(inarctan(2/x)) )] f^((n)) (x) = ((i(−1)^(n−1) (n−1)!)/(2(x^2 +4)^(n/2) ))[e^(inarctan(2/x)) −e^(−inarctan(2/x)) ] f^((n)) (x) = −(((−1)^(n−1) (n−1)!)/((x^2 +4)^(n/2) ))sin(narctan(2/x)) f^((n)) (x) = (((−1)^n (n−1)!)/((x^2 +4)^(n/2) ))sin(narctan(2/x)) f^((n)) (1) = (((−1)^n (n−1)!)/5^(n/2) )sin(narctan2) f^((7)) (x) = (((−1)^7 6!)/((x^2 +4)^(7/2) ))sin((1/7)arctan(2/x)) f^((7)) (x) = −((720)/((x^2 +4)^(7/2) ))sin((1/7)arctan(2/x)) f^((7)) ((1/7)) = −((720)/(((1/(49))+4)^(7/2) ))sin((1/7)arctan14) f^((7)) ((1/7)) = −((720)/((((197)/(49)))^(7/2) ))sin((1/7)arctan14) f^((7)) ((1/7)) = −((720×7^7 )/(197^3 (√(197))))sin((1/7)arctan14) f^((7)) ((1/7)) = −((823543)/( 38809(√(197))))sin((1/7)arctan14) ...maybe...

$$ \\ $$$${f}\left({x}\right)\:=\:\mathrm{arctan}\left(\frac{\mathrm{2}}{{x}}\right) \\ $$$${f}'\left({x}\right)\:=\:−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }}\:=\:−\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$${f}'\left({x}\right)\:=\:\frac{{i}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{x}−\mathrm{2}{i}}−\frac{\mathrm{1}}{{x}+\mathrm{2}{i}}\right] \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{{i}}{\mathrm{2}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\left[\frac{\mathrm{1}}{\left({x}−\mathrm{2}{i}\right)^{{n}} }−\frac{\mathrm{1}}{\left({x}+\mathrm{2}{i}\right)^{{n}} }\right] \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{{i}}{\mathrm{2}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\left[\frac{\mathrm{1}}{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}{e}^{−{i}\mathrm{arctan}\frac{\mathrm{2}}{{x}}} \right)^{{n}} }−\frac{\mathrm{1}}{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}{e}^{{i}\mathrm{arctan}\frac{\mathrm{2}}{{x}}} \right)^{{n}} }\right] \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{{i}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{{n}}{\mathrm{2}}} }\left[\frac{\mathrm{1}}{{e}^{−{in}\mathrm{arctan}\frac{\mathrm{2}}{{x}}} }−\frac{\mathrm{1}}{{e}^{{in}\mathrm{arctan}\frac{\mathrm{2}}{{x}}} }\right] \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{{i}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{{n}}{\mathrm{2}}} }\left[{e}^{{in}\mathrm{arctan}\frac{\mathrm{2}}{{x}}} −{e}^{−{in}\mathrm{arctan}\frac{\mathrm{2}}{{x}}} \right] \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:−\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{{n}}{\mathrm{2}}} }\mathrm{sin}\left({n}\mathrm{arctan}\frac{\mathrm{2}}{{x}}\right) \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{1}\right)!}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{{n}}{\mathrm{2}}} }\mathrm{sin}\left({n}\mathrm{arctan}\frac{\mathrm{2}}{{x}}\right) \\ $$$${f}^{\left({n}\right)} \left(\mathrm{1}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{1}\right)!}{\mathrm{5}^{\frac{{n}}{\mathrm{2}}} }\mathrm{sin}\left({n}\mathrm{arctan2}\right) \\ $$$${f}^{\left(\mathrm{7}\right)} \left({x}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{\mathrm{7}} \mathrm{6}!}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} }\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{7}}\mathrm{arctan}\frac{\mathrm{2}}{{x}}\right) \\ $$$${f}^{\left(\mathrm{7}\right)} \left({x}\right)\:=\:−\frac{\mathrm{720}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} }\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{7}}\mathrm{arctan}\frac{\mathrm{2}}{{x}}\right) \\ $$$${f}^{\left(\mathrm{7}\right)} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)\:=\:−\frac{\mathrm{720}}{\left(\frac{\mathrm{1}}{\mathrm{49}}+\mathrm{4}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} }\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{7}}\mathrm{arctan14}\right) \\ $$$${f}^{\left(\mathrm{7}\right)} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)\:=\:−\frac{\mathrm{720}}{\left(\frac{\mathrm{197}}{\mathrm{49}}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} }\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{7}}\mathrm{arctan14}\right) \\ $$$${f}^{\left(\mathrm{7}\right)} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)\:=\:−\frac{\mathrm{720}×\mathrm{7}^{\mathrm{7}} }{\mathrm{197}^{\mathrm{3}} \sqrt{\mathrm{197}}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{7}}\mathrm{arctan14}\right) \\ $$$${f}^{\left(\mathrm{7}\right)} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)\:=\:−\frac{\mathrm{823543}}{\:\mathrm{38809}\sqrt{\mathrm{197}}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{7}}\mathrm{arctan14}\right) \\ $$$$...\mathrm{maybe}... \\ $$

Commented by mathmax by abdo last updated on 30/Nov/20

thank you sir.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by mathmax by abdo last updated on 04/Dec/20

we have f(x)=arctan((2/x))⇒f^((1)) (x)=−(2/(x^2 (1+(4/x^2 ))))=((−2)/(x^2 +4)) =((−2)/((x−2i)(x+2i))) =((−2)/(4i)){(1/(x−2i))−(1/(x+2i))} =(1/(2i)){(1/(x+2i))−(1/(x−2i))} ⇒ f^((n)) (x)=(1/(2i)){((1/(x+2i)))^()n−1)) −((1/(x−2i)))^((n−1)) } =(1/(2i)){(((−1)^(n−1) (n−1)!)/((x+2i)^n ))−(((−1)^(n−1) (n−1)!)/((x−2i)^n ))} ⇒ f^((n)) (x)=(((−1)^(n−1) (n−1)!)/(2i)){(((x−2i)^n −(x+2i)^n )/((x^2 +4)^n ))} (n>0) f^((n)) (1) =(((−1)^(n−1) (n−1)!)/(2i)){(((1−2i)^n −(1+2i)^n )/5^n )} but 1−2i =(√5) e^(−iarctan(2)) and 1+2i =(√5)e^(iarctan(2)) ⇒ (1−2i)^n −(1+2i)^n =−(√5){ e^(iarctan(2)) −e^(−iarctan(2)) } =−(√5)(2isin(arctan(2))} ⇒ f^((n)) (1) =(((−1)^(n−1) (n−1)!)/(2i 5^n ))(−(√5)2isin(arctan(2)) =(1/5^(n−(1/2)) )(−1)^n (n−1)! sin(arctan2)

$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{arctan}\left(\frac{\mathrm{2}}{\mathrm{x}}\right)\Rightarrow\mathrm{f}^{\left(\mathrm{1}\right)} \left(\mathrm{x}\right)=−\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }\right)}=\frac{−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$$=\frac{−\mathrm{2}}{\left(\mathrm{x}−\mathrm{2i}\right)\left(\mathrm{x}+\mathrm{2i}\right)}\:=\frac{−\mathrm{2}}{\mathrm{4i}}\left\{\frac{\mathrm{1}}{\mathrm{x}−\mathrm{2i}}−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{2i}}\right\}\:=\frac{\mathrm{1}}{\mathrm{2i}}\left\{\frac{\mathrm{1}}{\mathrm{x}+\mathrm{2i}}−\frac{\mathrm{1}}{\mathrm{x}−\mathrm{2i}}\right\}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2i}}\left\{\left(\frac{\mathrm{1}}{\mathrm{x}+\mathrm{2i}}\right)^{\left.\right)\left.\mathrm{n}−\mathrm{1}\right)} −\left(\frac{\mathrm{1}}{\mathrm{x}−\mathrm{2i}}\right)^{\left(\mathrm{n}−\mathrm{1}\right)} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2i}}\left\{\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{x}+\mathrm{2i}\right)^{\mathrm{n}} }−\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\left(\mathrm{x}−\mathrm{2i}\right)^{\mathrm{n}} }\right\}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\mathrm{2i}}\left\{\frac{\left(\mathrm{x}−\mathrm{2i}\right)^{\mathrm{n}} −\left(\mathrm{x}+\mathrm{2i}\right)^{\mathrm{n}} }{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{n}} }\right\}\:\:\left(\mathrm{n}>\mathrm{0}\right) \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{1}\right)\:=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\mathrm{2i}}\left\{\frac{\left(\mathrm{1}−\mathrm{2i}\right)^{\mathrm{n}} −\left(\mathrm{1}+\mathrm{2i}\right)^{\mathrm{n}} }{\mathrm{5}^{\mathrm{n}} }\right\} \\ $$$$\mathrm{but}\:\mathrm{1}−\mathrm{2i}\:=\sqrt{\mathrm{5}}\:\mathrm{e}^{−\mathrm{iarctan}\left(\mathrm{2}\right)} \:\:\mathrm{and}\:\mathrm{1}+\mathrm{2i}\:=\sqrt{\mathrm{5}}\mathrm{e}^{\mathrm{iarctan}\left(\mathrm{2}\right)} \:\Rightarrow \\ $$$$\left(\mathrm{1}−\mathrm{2i}\right)^{\mathrm{n}} −\left(\mathrm{1}+\mathrm{2i}\right)^{\mathrm{n}} \:=−\sqrt{\mathrm{5}}\left\{\:\mathrm{e}^{\mathrm{iarctan}\left(\mathrm{2}\right)} −\mathrm{e}^{−\mathrm{iarctan}\left(\mathrm{2}\right)} \right\} \\ $$$$=−\sqrt{\mathrm{5}}\left(\mathrm{2isin}\left(\mathrm{arctan}\left(\mathrm{2}\right)\right)\right\}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{1}\right)\:=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!}{\mathrm{2i}\:\mathrm{5}^{\mathrm{n}} }\left(−\sqrt{\mathrm{5}}\mathrm{2isin}\left(\mathrm{arctan}\left(\mathrm{2}\right)\right)\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}} }\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{n}−\mathrm{1}\right)!\:\mathrm{sin}\left(\mathrm{arctan2}\right) \\ $$

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