derivative of ((x+1(2x-5))/(2x+1))

解答

\frac{d}{d x} (\frac{( x + 1 ) ( 2 x - 5 )}{2 x + 1})

\frac{4 x ^{2} + 4 x + 7}{( 2 x + 1 ) ^{2}}

求解步骤

\frac{d}{d x} (\frac{( x + 1 ) ( 2 x - 5 )}{2 x + 1})

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} ( ( x + 1 ) ( 2 x - 5 ) ) ( 2 x + 1 ) - \frac{d}{d x} ( 2 x + 1 ) ( x + 1 ) ( 2 x - 5 )}{( 2 x + 1 ) ^{2}}

\frac{d}{d x} ((x + 1) (2 x - 5)) = 4 x - 3

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

= \frac{( 4 x - 3 ) ( 2 x + 1 ) - 2 ( x + 1 ) ( 2 x - 5 )}{( 2 x + 1 ) ^{2}}

乘开

(4 x - 3) (2 x + 1) - 2 (x + 1) (2 x - 5) : 4 x^{2} + 4 x + 7

= \frac{4 x ^{2} + 4 x + 7}{( 2 x + 1 ) ^{2}}

\int \frac{x}{ln ( x )} d x

\int \frac{1}{5 - x} d x

a re a x^{2} + 2 x, x + 2

\int \frac{1}{tan ( x )} d x

\int_{0}^{0.5} 300 x d x