maclaurin e^{-5x}

Lösung

maclaurin

e^{- 5 x}

1 - 5 x + \frac{25}{2} x^{2} - \frac{125}{6} x^{3} + \frac{625}{24} x^{4} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( e ^{- 5 x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{- 5 x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{- 5 x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{- 5}{1 !} x + \frac{25}{2 !} x^{2} + \frac{- 125}{3 !} x^{3} + \frac{625}{4 !} x^{4} + \dots

Fasse zusammen

= 1 - 5 x + \frac{25}{2} x^{2} - \frac{125}{6} x^{3} + \frac{625}{24} x^{4} + \dots

3 y^{^{'}} - 6 y = 12

t \frac{d ^{2} y}{d t ^{2}} + 4 \frac{d y}{d t} = t^{2}

t an g e n t y = - \frac{3}{2} x^{2} + 9 x + 9, (- 1, - \frac{3}{2})

\int 7 - x^{2} d x

x \to - 2 lim (\frac{2 x}{2 x + 2})