maclaurin f(x)= 1/(2x-5)

Lösung

maclaurin

f (x) = \frac{1}{2 x - 5}

- \frac{1}{5} - \frac{2}{25} x - \frac{4}{125} x^{2} - \frac{8}{625} x^{3} - \frac{16}{3125} x^{4} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

= - \frac{1}{5} + \frac{\frac{d}{d x} ( \frac{1}{2 x - 5} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{2 x - 5} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{2 x - 5} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= - \frac{1}{5} + \frac{- \frac{2}{25}}{1 !} x + \frac{- \frac{8}{125}}{2 !} x^{2} + \frac{- \frac{48}{625}}{3 !} x^{3} + \frac{- \frac{384}{3125}}{4 !} x^{4} + \dots

Fasse zusammen

= - \frac{1}{5} - \frac{2}{25} x - \frac{4}{125} x^{2} - \frac{8}{625} x^{3} - \frac{16}{3125} x^{4} + \dots

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

d er i v a t i v e 3 x + 5

\frac{d y}{d t} = 4 y (1 - y)

\int_{1}^{2} \frac{2}{x ^{3} + x} d x

d er i v a t i v e 4 x^{5} + 6 x