maclaurin 8/(3x+3)

Lösung

maclaurin

\frac{8}{3 x + 3}

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

Schritte zur Lösung

Apply the Maclaurin Formula

= \frac{8}{3} + \frac{\frac{d}{d x} ( \frac{8}{3 x + 3} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{8}{3 x + 3} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{8}{3 x + 3} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= \frac{8}{3} + \frac{- \frac{8}{3}}{1 !} x + \frac{\frac{16}{3}}{2 !} x^{2} + \frac{- 16}{3 !} x^{3} + \frac{64}{4 !} x^{4} + \dots

Fasse zusammen

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

d er i v a t i v e f (x) = \frac{6 cos ( x )}{7 - 9 cos ( x )}

d er i v a t i v e cos (2 x^{2})

\int_{0}^{2} \frac{3 ln ( x )}{x} d x

\frac{d}{d x} (\frac{1}{4} e^{x})

\frac{d}{d x} (ln (\frac{3 + x}{3 - x}))