derivative f(x)=(x^2-5x)/(x+1)

解答

导数

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

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

求解步骤

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

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

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

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

化简

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

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

\int 3 x^{2} x^{3} - 5 d x

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

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