maclaurin (x+1)ln(x+1)

Lösung

maclaurin

(x + 1) ln (x + 1)

x + \frac{1}{2} x^{2} - \frac{1}{6} x^{3} + \frac{1}{12} x^{4} - \frac{1}{20} x^{5} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{1}{2 !} x^{2} + \frac{- 1}{3 !} x^{3} + \frac{2}{4 !} x^{4} + \frac{- 6}{5 !} x^{5} + \dots

Fasse zusammen

= x + \frac{1}{2} x^{2} - \frac{1}{6} x^{3} + \frac{1}{12} x^{4} - \frac{1}{20} x^{5} + \dots

t an g e n t y = (\frac{4}{x + 2}), at x = 2

\int \frac{6}{7} sec^{2} (\frac{x}{7}) d x

\int \frac{e ^{3 - x}}{( 1 + e ^{3 - x} ) ^{2}} d x

x \to 0 - lim (\frac{x}{13 x ^{4} - 6 x})

\frac{\partial}{\partial y} (x yz)