derivative of x/(x^{-1+9})

解答

\frac{d}{d x} (\frac{x}{x ^{- 1} + 9})

\frac{x ( 9 x + 2 )}{( 1 + 9 x ) ^{2}}

求解步骤

\frac{d}{d x} (\frac{x}{x ^{- 1} + 9})

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} ( x ^{- 1} + 9 ) - \frac{d}{d x} ( x ^{- 1} + 9 ) x}{( x ^{- 1} + 9 ) ^{2}}

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

\frac{d}{d x} (x^{- 1} + 9) = - \frac{1}{x ^{2}}

= \frac{1 \cdot ( x ^{- 1} + 9 ) - ( - \frac{1}{x ^{2}} ) x}{( x ^{- 1} + 9 ) ^{2}}

化简

\frac{1 \cdot ( x ^{- 1} + 9 ) - ( - \frac{1}{x ^{2}} ) x}{( x ^{- 1} + 9 ) ^{2}} : \frac{x ( 9 x + 2 )}{( 1 + 9 x ) ^{2}}

= \frac{x ( 9 x + 2 )}{( 1 + 9 x ) ^{2}}

\int (\frac{3 x ^{2}}{64 - x ^{2}}) d x

x \to 0 + lim (\frac{x ^{- 2}}{x})

\int x 36 - x^{2} d x

\int \frac{e ^{t}}{( 1 + e ^{(2 t)} ) ^{(\frac{3}{2})}} d t

\frac{d y}{d x} = 3 y \cdot e^{x + 3}