derivative f(x)=(e^xx-e^x)/(x^2)

解答

导数

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

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

求解步骤

\frac{d}{d x} (\frac{e ^{x} x - e ^{x}}{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} ( e ^{x} x - e ^{x} ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( e ^{x} x - e ^{x} )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (e^{x} x - e^{x}) = e^{x} x

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

= \frac{e ^{x} x x ^{2} - 2 x ( e ^{x} x - e ^{x} )}{( x ^{2} ) ^{2}}

化简

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

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

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

ma c l a u r in f (x) = \frac{1}{( 1 + x )} 1

x^{^{'}} - a x - 1 = 0

\int_{0}^{1} x \cdot 3^{x^{2} + 1} d x

y^{^{''}} + 9 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 0