maclaurin ln(1+4x^2+4x)

Lösung

maclaurin

ln (1 + 4 x^{2} + 4 x)

4 x - 4 x^{2} + \frac{16}{3} x^{3} - 8 x^{4} + \frac{64}{5} x^{5} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{4}{1 !} x + \frac{- 8}{2 !} x^{2} + \frac{32}{3 !} x^{3} + \frac{- 192}{4 !} x^{4} + \frac{1536}{5 !} x^{5} + \dots

Fasse zusammen

= 4 x - 4 x^{2} + \frac{16}{3} x^{3} - 8 x^{4} + \frac{64}{5} x^{5} + \dots

\frac{\partial}{\partial y} (e^{x y z^{5}} y z^{5})

\frac{d}{d x} (x + cos (x^{2}))

\int 5 e^{3 x} d x

e x t re m e h (w) = \frac{( x + 1 ) e ^{x}}{w - 2}

d er i v a t i v e f (x) = 5 x^{2} - 5 x