derivative of x/(1-e^t(1-x))

解答

\frac{d}{d x} (\frac{x}{1 - e ^{t} ( 1 - x )})

\frac{1 - e ^{t}}{( 1 - e ^{t} ( 1 - x ) ) ^{2}}

求解步骤

\frac{d}{d x} (\frac{x}{1 - e ^{t} ( 1 - x )})

将

t

视为常数

使用微分除法定则:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

= \frac{1 \cdot ( 1 - e ^{t} ( 1 - x ) ) - e ^{t} x}{( 1 - e ^{t} ( 1 - x ) ) ^{2}}

化简

\frac{1 \cdot ( 1 - e ^{t} ( 1 - x ) ) - e ^{t} x}{( 1 - e ^{t} ( 1 - x ) ) ^{2}} : \frac{1 - e ^{t}}{( 1 - e ^{t} ( 1 - x ) ) ^{2}}

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

e x p an d \frac{[ 100 \cdot ( s + 1 ) ]}{[ ( s ^{2} ) \cdot ( s + 10 ) ]}

e x p an d (2 i, 3 j, - 4 k) (i - 3 j + k)

e x p an d \frac{8 ( 8 - x )}{64 - x}

e x p an d (- 10 + 11) (- 4 + 10)

e x p an d \frac{( ( x + y ) ( ( x + y ) + 1 ) - ( x x + 1 ) )}{y}