taylor 2/((3x+1)),x=2

Lösung

taylor

\frac{2}{( 3 x + 1 )}, x = 2

\frac{2}{7} - \frac{6}{49} (x - 2) + \frac{18}{343} (x - 2)^{2} - \frac{54}{2401} (x - 2)^{3} + \frac{162}{16807} (x - 2)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= \frac{2}{7} + \frac{\frac{d}{d x} ( \frac{2}{3 x + 1} ) ( 2 )}{1 !} (x - 2) + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{2}{3 x + 1} ) ( 2 )}{2 !} (x - 2)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{2}{3 x + 1} ) ( 2 )}{3 !} (x - 2)^{3} + \dots

Evaluate Derivatives

= \frac{2}{7} + \frac{- \frac{6}{49}}{1 !} (x - 2) + \frac{\frac{36}{343}}{2 !} (x - 2)^{2} + \frac{- \frac{324}{2401}}{3 !} (x - 2)^{3} + \frac{\frac{3888}{16807}}{4 !} (x - 2)^{4} + \dots

Fasse zusammen

= \frac{2}{7} - \frac{6}{49} (x - 2) + \frac{18}{343} (x - 2)^{2} - \frac{54}{2401} (x - 2)^{3} + \frac{162}{16807} (x - 2)^{4} + \dots

\int - x e^{- l n (x)} d x

t an g e n t f (x) = 2 x^{3} + 3 x^{2}, at x = 1

\frac{d}{d x} (25 (x^{2} - y^{2}))

\frac{d}{d x} (\frac{1 0 ^{x}}{1 0 ^{x} + 4})

\frac{d}{d x} (e^{- 6 x^{2}})