maclaurin e^{-3x}

Lösung

maclaurin

e^{- 3 x}

1 - 3 x + \frac{9}{2} x^{2} - \frac{9}{2} x^{3} + \frac{27}{8} x^{4} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{- 3}{1 !} x + \frac{9}{2 !} x^{2} + \frac{- 27}{3 !} x^{3} + \frac{81}{4 !} x^{4} + \dots

Fasse zusammen

= 1 - 3 x + \frac{9}{2} x^{2} - \frac{9}{2} x^{3} + \frac{27}{8} x^{4} + \dots

(f (x) (x^{2}))^{'}

f (x) = \frac{sin ( x )}{cos ( ln ( - x ) )}

s l o p e x y - 4 y^{2} = 8

\int 5 x^{3} d x

\int \frac{( 2 t ^{2} + t t - 1 )}{t ^{2}} d t