let f(x)=arctan(2x−(1/x)) find f^((n)) (x) andf^((n)) (1)


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Question Number 83254 by mathmax by abdo last updated on 29/Feb/20

$l e t f (x) = a r c t a n (2 x - \frac{1}{x})$ $f i n d f^{(n)} (x) {a n d f}^{(n)} (1)$
Commented by mathmax by abdo last updated on 02/Mar/20
$we have f^′ (x)=((2+(1/x^2 ))/(1+(2x−(1/x))^2 )) =((2x^2 +1)/(x^2 {1+(((2x^2 −1)^2 )/x^2 )})) =((2x^2 +1)/(x^2 +(2x^2 −1)^2 )) =((2x^2 +1)/(x^2 +4x^4 −4x^2 +1)) =((2x^2 +1)/(4x^4 −3x^2 +1)) 4x^4 −3x^2 +1=0 →4t^2 −3t +1=0 (t=x^2 ) Δ=9−16 =−7 ⇒t_1 =((3+i(√7))/8) and t_2 =((3−i(√7))/8) ⇒ f^′ (x)=((2x^2 +1)/(4(x^2 −t_1 )(x^2 −t_2 ))) =(1/(4(t_1 −t_2 )))((1/(x^2 −t_1 ))−(1/(x^2 −t_2 )))(2x^2 +1) =(1/(i(√7))){ ((2x^2 +1)/(x^2 −t_1 ))−((2x^2 +1)/(x^2 −t_2 ))} =(1/(i(√7))){((2(x^2 −t_1 )+1+2t_1 )/(x^2 −t_1 ))−((2(x^2 −t_2 )+1+2t_2 )/(x^2 −t_2 ))} =(1/(i(√7))){((2t_1 +1)/(x^2 −t_1 ))−((2t_2 +1)/(x^2 −t_2 ))} =(1/(i(√7))){((2t_1 +1)/(2(√t_1 )))×((1/(x−(√t_1 )))−(1/(x+(√t_1 )))) −((2t_2 +1)/(2(√t_2 )))((1/(x−(√t_2 )))−(1/(x+(√t_2 ))))}⇒ f^((n)) (x) =((2t_1 +1)/(2i(√7)(√t_1 ))){ (((−1)^(n−1) (n−1)!)/((x−(√t_1 ))^n ))−(((−1)^(n−1) (n−1)!)/((x+(√t_1 ))^n ))} −((2t_2 +1)/(2i(√7)(√t_2 ))){ (((−1)^(n−1) (n−1)!)/((x−(√t_2 ))^n ))−(((−1)^(n−1) (n−1)!)/((x+(√t_2 ))^n ))} be continued..$
$w e h a v e f^{^{'}} (x) = \frac{2 + \frac{1}{x^{2}}}{1 + {(2 x - \frac{1}{x})}^{2}} = \frac{2 x^{2} + 1}{x^{2} {1 + \frac{{(2 x^{2} - 1)}^{2}}{x^{2}}}}$ $= \frac{2 x^{2} + 1}{x^{2} + {(2 x^{2} - 1)}^{2}} = \frac{2 x^{2} + 1}{x^{2} + 4 x^{4} - 4 x^{2} + 1} = \frac{2 x^{2} + 1}{4 x^{4} - 3 x^{2} + 1}$ $4 x^{4} - 3 x^{2} + 1 = 0 \to 4 t^{2} - 3 t + 1 = 0 (t = x^{2})$ $Δ = 9 - 16 = - 7 \Rightarrow t_{1} = \frac{3 + i \sqrt{7}}{8} a n d t_{2} = \frac{3 - i \sqrt{7}}{8} \Rightarrow$ $f^{^{'}} (x) = \frac{2 x^{2} + 1}{4 (x^{2} - t_{1}) (x^{2} - t_{2})} = \frac{1}{4 (t_{1} - t_{2})} (\frac{1}{x^{2} - t_{1}} - \frac{1}{x^{2} - t_{2}}) (2 x^{2} + 1)$ $= \frac{1}{i \sqrt{7}} {\frac{2 x^{2} + 1}{x^{2} - t_{1}} - \frac{2 x^{2} + 1}{x^{2} - t_{2}}} = \frac{1}{i \sqrt{7}} {\frac{2 (x^{2} - t_{1}) + 1 + 2 t_{1}}{x^{2} - t_{1}} - \frac{2 (x^{2} - t_{2}) + 1 + 2 t_{2}}{x^{2} - t_{2}}}$ $= \frac{1}{i \sqrt{7}} {\frac{2 t_{1} + 1}{x^{2} - t_{1}} - \frac{2 t_{2} + 1}{x^{2} - t_{2}}}$ $= \frac{1}{i \sqrt{7}} {\frac{2 t_{1} + 1}{2 \sqrt{t_{1}}} \times (\frac{1}{x - \sqrt{t_{1}}} - \frac{1}{x + \sqrt{t_{1}}}) - \frac{2 t_{2} + 1}{2 \sqrt{t_{2}}} (\frac{1}{x - \sqrt{t_{2}}} - \frac{1}{x + \sqrt{t_{2}}})} \Rightarrow$ $f^{(n)} (x) = \frac{2 t_{1} + 1}{2 i \sqrt{7} \sqrt{t_{1}}} {\frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - \sqrt{t_{1}})}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + \sqrt{t_{1}})}^{n}}}$ $- \frac{2 t_{2} + 1}{2 i \sqrt{7} \sqrt{t_{2}}} {\frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - \sqrt{t_{2}})}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + \sqrt{t_{2}})}^{n}}} b e c o n t i n u e d . .$
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