derivative f(x)=(2x^2-x-1)/(2x+2)

解答

导数

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

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

求解步骤

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

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

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

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

化简

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

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

x \to 0 + lim (x^{20 x})

\frac{d}{d x} (7 x^{4} cot (x))

x y d x - (1 + x^{2}) d y = 0

x^{''}

\frac{d y}{d x} + 3 y = 40