maclaurin f(x)=arctan(2x)

Lösung

maclaurin

f (x) = arctan (2 x)

2 x - \frac{8}{3} x^{3} + \frac{32}{5} x^{5} - \frac{128}{7} x^{7} + \frac{512}{9} x^{9} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d x} ( arctan ( 2 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( arctan ( 2 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( arctan ( 2 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{2}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 16}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{768}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 92160}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{20643840}{9 !} x^{9} + \dots

Fasse zusammen

= 2 x - \frac{8}{3} x^{3} + \frac{32}{5} x^{5} - \frac{128}{7} x^{7} + \frac{512}{9} x^{9} + \dots

\int (2 x - 6)^{3} d x

\int_{0}^{3} (x + 1)^{\frac{1}{2}} d x

y^{^{'}} - y = e^{- t}

\frac{\partial}{\partial x} (x e^{(\frac{y}{z})})

\frac{\partial}{\partial x} (\frac{n}{x + 1})