taylor sin(2x),0

Lösung

taylor

sin (2 x), 0

2 x - \frac{4}{3} x^{3} + \frac{4}{15} x^{5} - \frac{8}{315} x^{7} + \frac{4}{2835} x^{9} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{2}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 8}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{32}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 128}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{512}{9 !} x^{9} + \dots

Fasse zusammen

= 2 x - \frac{4}{3} x^{3} + \frac{4}{15} x^{5} - \frac{8}{315} x^{7} + \frac{4}{2835} x^{9} + \dots

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

d er i v a t i v e arcsec (x) \cdot arccos (x)

\int t e^{t + 2} d t

\int \frac{d m}{m ^{3} + m ^{2}} d x

\frac{d}{d r} (\frac{1}{2 r} + \frac{1}{7 r ^{\frac{6}{7}}})