taylor x^3-4sqrt(x)+log_{e}(x)

Lösung

taylor

x^{3} - 4 x + lo g_{e} (x)

- 3 + 2 (x - 1) + 3 (x - 1)^{2} + \frac{13}{12} (x - 1)^{3} - \frac{3}{32} (x - 1)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= - 3 + \frac{\frac{d}{d x} ( x ^{3} - 4 x + lo g _{e} ( x ) ) ( 1 )}{1 !} (x - 1) + \frac{\frac{d ^{2}}{d x ^{2}} ( x ^{3} - 4 x + lo g _{e} ( x ) ) ( 1 )}{2 !} (x - 1)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ^{3} - 4 x + lo g _{e} ( x ) ) ( 1 )}{3 !} (x - 1)^{3} + \dots

Evaluate Derivatives

= - 3 + \frac{2}{1 !} (x - 1) + \frac{6}{2 !} (x - 1)^{2} + \frac{\frac{13}{2}}{3 !} (x - 1)^{3} + \frac{- \frac{9}{4}}{4 !} (x - 1)^{4} + \dots

Fasse zusammen

= - 3 + 2 (x - 1) + 3 (x - 1)^{2} + \frac{13}{12} (x - 1)^{3} - \frac{3}{32} (x - 1)^{4} + \dots

\frac{\partial}{\partial x} (- 3 x y^{2} + 2 y cos (x))

\frac{\partial}{\partial y} (2 (y + 1) x^{2})

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

\int 0.06 q - 0.6 q - \frac{7700}{q ^{2}} d q

\frac{d y}{d x} = a - b y^{2}