((2-2x)/(2x-x^2+3))^'

解答

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

\frac{- 2 x ^{2} + 4 x - 10}{( 2 x - x ^{2} + 3 ) ^{2}}

求解步骤

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

使用微分除法定则:

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

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

(2 - 2 x)^{'} = - 2

(2 x - x^{2} + 3)^{'} = - 2 x + 2

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

化简

\frac{( - 2 ) ( 2 x - x ^{2} + 3 ) - ( - 2 x + 2 ) ( 2 - 2 x )}{( 2 x - x ^{2} + 3 ) ^{2}} : \frac{- 2 x ^{2} + 4 x - 10}{( 2 x - x ^{2} + 3 ) ^{2}}

= \frac{- 2 x ^{2} + 4 x - 10}{( 2 x - x ^{2} + 3 ) ^{2}}

\int \frac{1}{( z ^{2} + x ^{2} ) ^{\frac{3}{2}}} d x

\int \frac{100}{x ^{2}} d x

d er i v a t i v e x^{2} + 3 x + 14

\int \frac{x}{8 + x ^{2}} d x

a re a y = x, y = x^{4}, (0, 1)