maclaurin e^{3x^3}

Lösung

maclaurin

e^{3 x^{3}}

1 + 3 x^{3} + \frac{9}{2} x^{6} + \frac{9}{2} x^{9} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{0}{2 !} x^{2} + \frac{18}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{3240}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{1632960}{9 !} x^{9} + \frac{0}{10 !} x^{10} + \dots

Fasse zusammen

= 1 + 3 x^{3} + \frac{9}{2} x^{6} + \frac{9}{2} x^{9} + \dots

\int \frac{1}{( 4 - x ^{2} ) ^{2}} d x

\frac{\partial}{\partial x} (5 ln (1 + x^{2} - y))

x \to 0 lim (\frac{ln ( x )}{x ^{2}})

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

f^{^{'}} (x) = 5 (1 + e^{x})^{4} e^{x}