taylor ln(-2x-5),-5

Lösung

taylor

ln (- 2 x - 5), - 5

ln (5) - \frac{2}{5} (x + 5) - \frac{2}{25} (x + 5)^{2} - \frac{8}{375} (x + 5)^{3} - \frac{4}{625} (x + 5)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= ln (5) + \frac{- \frac{2}{5}}{1 !} (x + 5) + \frac{- \frac{4}{25}}{2 !} (x + 5)^{2} + \frac{- \frac{16}{125}}{3 !} (x + 5)^{3} + \frac{- \frac{96}{625}}{4 !} (x + 5)^{4} + \dots

Fasse zusammen

= ln (5) - \frac{2}{5} (x + 5) - \frac{2}{25} (x + 5)^{2} - \frac{8}{375} (x + 5)^{3} - \frac{4}{625} (x + 5)^{4} + \dots

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

\int (8 x + e^{x}) 4 x^{2} + e^{x} d x

\frac{d ^{2}}{d x ^{2}} (3 x^{2} cos (8 x))

x \to 1 lim ((ln (x))^{x - 1})

\frac{d y}{d x} = 9 x^{2} - 12 x