taylor ln(4x),1

Lösung

taylor

ln (4 x), 1

2 ln (2) + (x - 1) - \frac{1}{2} (x - 1)^{2} + \frac{1}{3} (x - 1)^{3} - \frac{1}{4} (x - 1)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

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

\int \frac{2 - x}{2 + x} d x

\frac{\partial}{\partial x} (x y - x sin (y))

\frac{( t ^{2} + 1 ) ^{\frac{3}{2}}}{3}^{^{'}}

\frac{d}{d x} ((x + 2) (x - 2)^{3})

d er i v a t i v e - 7 (6 x^{2} + 7)^{- 6}