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

解答

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

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

求解步骤

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

使用微分除法定则:

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

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

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

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

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

化简

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

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

\frac{\partial}{\partial x} (- 4 y (- 2 x - y - 3))

\int 1 - \frac{16}{sin ^{4} ( x )} d x

x \to 0 lim (x ln^{2} (x))

\frac{\partial}{\partial x} (3 y^{2} - x^{2})

\frac{d}{d x} (\frac{x ^{2} + 4}{4 x ^{3} - 5 x})