maclaurin (1+x)e^{-x}

Lösung

maclaurin

(1 + x) e^{- x}

1 - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \frac{1}{8} x^{4} + \frac{1}{30} x^{5} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 1}{2 !} x^{2} + \frac{2}{3 !} x^{3} + \frac{- 3}{4 !} x^{4} + \frac{4}{5 !} x^{5} + \dots

Fasse zusammen

= 1 - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \frac{1}{8} x^{4} + \frac{1}{30} x^{5} + \dots

d er i v a t i v e \frac{e ^{x} - e ^{- x}}{2}

\frac{d}{d x} ((x)^{\frac{1}{3}})

\frac{d}{d x} (- \frac{ln ( 1 + 6 x )}{x})

\frac{d}{d x} (\frac{1}{x ( x - 1 )})

x \frac{d y}{d x} = 6 y