麦克劳林 1/(\sqrt[3]{2x+1)}

解答

麦克劳林

\frac{1}{3 2 x + 1}

1 - \frac{2}{3} x + \frac{8}{9} x^{2} - \frac{112}{81} x^{3} + \frac{560}{243} x^{4} + \dots

求解步骤

应用麦克劳林公式

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

求导数

= 1 + \frac{- \frac{2}{3}}{1 !} x + \frac{\frac{16}{9}}{2 !} x^{2} + \frac{- \frac{224}{27}}{3 !} x^{3} + \frac{\frac{4480}{81}}{4 !} x^{4} + \dots

整理后得

= 1 - \frac{2}{3} x + \frac{8}{9} x^{2} - \frac{112}{81} x^{3} + \frac{560}{243} x^{4} + \dots

\frac{d ^{2}}{d x ^{2}} (e^{5 x})

\frac{\partial}{\partial x} (10 x y)

d er i v a t i v e y = 2 x^{2} e^{x} + 4 x^{3}

d er i v a t i v e \frac{x}{x + 13}

x \to - 5 lim (\frac{x ^{2} + 6}{x + 2})