taylor ln(1-5y)

Lösung

taylor

ln (1 - 5 y)

- 5 y - \frac{25}{2} y^{2} - \frac{125}{3} y^{3} - \frac{625}{4} y^{4} - 625 y^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{- 5}{1 !} y + \frac{- 25}{2 !} y^{2} + \frac{- 250}{3 !} y^{3} + \frac{- 3750}{4 !} y^{4} + \frac{- 75000}{5 !} y^{5} + \dots

Fasse zusammen

= - 5 y - \frac{25}{2} y^{2} - \frac{125}{3} y^{3} - \frac{625}{4} y^{4} - 625 y^{5} + \dots

in v erse l a pl a ce \frac{1}{s ^{2} + 9 s + 9}

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\frac{\partial}{\partial x} (\frac{( a x + b y )}{c x + d y})

x y^{^{''}} + y^{^{'}} = 1

\frac{\partial}{\partial x} (6 x sin (x + 2 y - z))