taylor 2xe^{-4x}

Lösung

taylor

2 x e^{- 4 x}

2 x - 8 x^{2} + 16 x^{3} - \frac{64}{3} x^{4} + \frac{64}{3} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( 2 x e ^{- 4 x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 2 x e ^{- 4 x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 2 x e ^{- 4 x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{2}{1 !} x + \frac{- 16}{2 !} x^{2} + \frac{96}{3 !} x^{3} + \frac{- 512}{4 !} x^{4} + \frac{2560}{5 !} x^{5} + \dots

Fasse zusammen

= 2 x - 8 x^{2} + 16 x^{3} - \frac{64}{3} x^{4} + \frac{64}{3} x^{5} + \dots

\int 1 - \frac{1}{x ^{2}} d x

\int 7 x^{2} e^{6 x} d x

t an g e n t f (x) = (x + 2 sin (x))

\int \frac{x - 3}{3 - 4 x ^{2}} d x

\frac{d y}{d x} = 8 x y