derivative of (1/(x^3-2/x)(x^2+x))

解答

\frac{d}{d x} ((\frac{1}{x ^{3}} - \frac{2}{x}) (x^{2} + x))

- \frac{1}{x ^{2}} - \frac{2}{x ^{3}} - 2

求解步骤

\frac{d}{d x} ((\frac{1}{x ^{3}} - \frac{2}{x}) (x^{2} + x))

化简

(\frac{1}{x ^{3}} - \frac{2}{x}) (x^{2} + x) : \frac{1}{x} + \frac{1}{x ^{2}} - 2 x - 2

= \frac{d}{d x} (\frac{1}{x} + \frac{1}{x ^{2}} - 2 x - 2)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{1}{x}) + \frac{d}{d x} (\frac{1}{x ^{2}}) - \frac{d}{d x} (2 x) - \frac{d}{d x} (2)

\frac{d}{d x} (\frac{1}{x}) = - \frac{1}{x ^{2}}

\frac{d}{d x} (\frac{1}{x ^{2}}) = - \frac{2}{x ^{3}}

\frac{d}{d x} (2 x) = 2

\frac{d}{d x} (2) = 0

= - \frac{1}{x ^{2}} - \frac{2}{x ^{3}} - 2 - 0

化简

= - \frac{1}{x ^{2}} - \frac{2}{x ^{3}} - 2

\int - 2 sin (n x) d x

t an g e n t 4 x^{2} = ln (x - y) + 3, at x = 0

\int e^{x} (4 + e^{x})^{5} d x

d e g 2 r a d (x^{''} + x)^{^{'}}

x \frac{d y}{d x} + y = e^{x}