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

\int sin (ln (3 x)) d x

\frac{d}{d x} (\frac{4}{x ^{2} + 2})

\int_{0}^{l n (2)} e^{2 x} d x

\int \frac{9}{2} (t + 1)^{\frac{1}{2}} d t

l a pl a ce t r an s f or m e^{- 1}