((x^2+x+4)/(x+1))^'

解答

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

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

求解步骤

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

使用微分除法定则:

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

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

(x^{2} + x + 4)^{'} = 2 x + 1

(x + 1)^{'} = 1

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

化简

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

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

l a pl a ce t r an s f or m 8 t e^{2 t}

d er i v a t i v e f (x) = \frac{5 - x e ^{x}}{x + e ^{x}}

d er i v a t i v e \frac{1}{2} sin^{2} (x)

x \to - 1 lim ((1 + \frac{1}{x})^{x})

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