let-f-x-arctan-2x-1-x-find-f-n-x-andf-n-1-

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

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 . .