taylor f(x)=(1+x)^{-4}

Lösung

taylor

f (x) = (1 + x)^{- 4}

1 - 4 x + 10 x^{2} - 20 x^{3} + 35 x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( ( 1 + x ) ^{- 4} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ( 1 + x ) ^{- 4} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ( 1 + x ) ^{- 4} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{- 4}{1 !} x + \frac{20}{2 !} x^{2} + \frac{- 120}{3 !} x^{3} + \frac{840}{4 !} x^{4} + \dots

Fasse zusammen

= 1 - 4 x + 10 x^{2} - 20 x^{3} + 35 x^{4} + \dots

d er i v a t i v e \frac{x}{( \frac{x}{10} ) ^{^{'}}}

s l o p e x^{2} - y^{2} = 64, (8, 0)

a re a 2 x - x^{2}, 8 - 4 x, 4 x

\frac{\partial}{\partial t} (- 3 s^{2} t^{2})

\frac{d}{d x} ((10 x^{2} - 20 x + 20) e^{x})