taylor e^{9x}

Lösung

taylor

e^{9 x}

1 + 9 x + \frac{81}{2} x^{2} + \frac{243}{2} x^{3} + \frac{2187}{8} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( e ^{9 x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{9 x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{9 x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{9}{1 !} x + \frac{81}{2 !} x^{2} + \frac{729}{3 !} x^{3} + \frac{6561}{4 !} x^{4} + \dots

Fasse zusammen

= 1 + 9 x + \frac{81}{2} x^{2} + \frac{243}{2} x^{3} + \frac{2187}{8} x^{4} + \dots

d er i v a t i v e f (t) = sin^{2} (t)

d er i v a t i v e \frac{x ^{3}}{2}

\int 8 d x

\int 50 x^{9} e^{5 x^{5}} d x

\frac{d}{d x} (\frac{x ^{2}}{sin ^{2} ( x )})