derivative of (-2/(x^4)+1/(x^3)-x)

解答

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

\frac{8}{x ^{5}} - \frac{3}{x ^{4}} - 1

求解步骤

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

化简

\frac{- 2}{x ^{4}} + \frac{1}{x ^{3}} - x : - \frac{2}{x ^{4}} + \frac{1}{x ^{3}} - x

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

使用微分加减法定则:

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

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

\frac{d}{d x} (\frac{2}{x ^{4}}) = - \frac{8}{x ^{5}}

\frac{d}{d x} (\frac{1}{x ^{3}}) = - \frac{3}{x ^{4}}

\frac{d x}{d x} = 1

= - (- \frac{8}{x ^{5}}) - \frac{3}{x ^{4}} - 1

化简

= \frac{8}{x ^{5}} - \frac{3}{x ^{4}} - 1

\frac{\partial}{\partial t} (5 t)

d er i v a t i v e x^{2} + 2

l c m 4 cos^{2} (x), \frac{sec ^{2} ( x )}{4}

\int 40 x e^{x} d x

d er i v a t i v e f (t) = (1 + t) (6 t^{2} - 9)