taylor 2xln(x+1)

Lösung

taylor

2 x ln (x + 1)

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

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{4}{2 !} x^{2} + \frac{- 6}{3 !} x^{3} + \frac{16}{4 !} x^{4} + \frac{- 60}{5 !} x^{5} + \frac{288}{6 !} x^{6} + \dots

Fasse zusammen

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

d er i v a t i v e y = \frac{200}{0.9 + 0.1 e ^{2 x}}

x \to 0 lim (\frac{2}{x ^{2} + x})

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

\frac{\partial}{\partial v} (2 v cos (u))

d er i v a t i v e 0.882 A^{0.842}