taylor 9xln(1+2x)

Lösung

taylor

9 x ln (1 + 2 x)

18 x^{2} - 18 x^{3} + 24 x^{4} - 36 x^{5} + \frac{288}{5} x^{6} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( 9 x ln ( 1 + 2 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 9 x ln ( 1 + 2 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 9 x ln ( 1 + 2 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{36}{2 !} x^{2} + \frac{- 108}{3 !} x^{3} + \frac{576}{4 !} x^{4} + \frac{- 4320}{5 !} x^{5} + \frac{41472}{6 !} x^{6} + \dots

Fasse zusammen

= 18 x^{2} - 18 x^{3} + 24 x^{4} - 36 x^{5} + \frac{288}{5} x^{6} + \dots

\frac{d}{d x} (ln (x + y^{2}))

x \frac{d y}{d x} = y^{2} - y, y (\frac{1}{4}) = \frac{1}{4}

\frac{\partial}{\partial x} (\frac{e ^{x}}{y + x ^{7}})

t an g e n t 4 x^{3} - 18 x^{2} - 48 x

\frac{d}{d x} (4 - 2 x)