(\partial)/(\partial x)(1/x-y/x)

解答

\frac{\partial}{\partial x} (\frac{1}{x} - \frac{y}{x})

\frac{- 1 + y}{x ^{2}}

求解步骤

\frac{\partial}{\partial x} (\frac{1}{x} - \frac{y}{x})

将

y

视为常数

使用微分加减法定则:

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

= \frac{\partial}{\partial x} (\frac{1}{x}) - \frac{\partial}{\partial x} (\frac{y}{x})

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

\frac{\partial}{\partial x} (\frac{y}{x}) = - \frac{y}{x ^{2}}

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

化简

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

= \frac{- 1 + y}{x ^{2}}

\int tan^{2} (x) tan (x) d x

\int \frac{7}{( x - 3 ) x ^{2} - 6 x + 5} d x

\frac{\partial}{\partial x} (e^{- x} sin (x))

\frac{d}{d x} (\frac{8}{x - 5})

f (x) = - csc^{2} (x)