maclaurin 1/(4z-5)

Lösung

maclaurin

\frac{1}{4 z - 5}

- \frac{1}{5} - \frac{4}{25} z - \frac{16}{125} z^{2} - \frac{64}{625} z^{3} - \frac{256}{3125} z^{4} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

= - \frac{1}{5} + \frac{\frac{d}{d z} ( \frac{1}{4 z - 5} ) ( 0 )}{1 !} z + \frac{\frac{d ^{2}}{d z ^{2}} ( \frac{1}{4 z - 5} ) ( 0 )}{2 !} z^{2} + \frac{\frac{d ^{3}}{d z ^{3}} ( \frac{1}{4 z - 5} ) ( 0 )}{3 !} z^{3} + \dots

Evaluate Derivatives

= - \frac{1}{5} + \frac{- \frac{4}{25}}{1 !} z + \frac{- \frac{32}{125}}{2 !} z^{2} + \frac{- \frac{384}{625}}{3 !} z^{3} + \frac{- \frac{6144}{3125}}{4 !} z^{4} + \dots

Fasse zusammen

= - \frac{1}{5} - \frac{4}{25} z - \frac{16}{125} z^{2} - \frac{64}{625} z^{3} - \frac{256}{3125} z^{4} + \dots

t an g e n t y = (x^{2} - 35)^{6}, at x = 6

d er i v a t i v e f (x) = \frac{x}{x ^{2} - 2}

d er i v a t i v e 3 x \cdot y

d er i v a t i v e f (x) = 2 x^{3}

\int_{0}^{\frac{π}{4}} 5 tan^{3} (x) d x