derivative of (-e^xx+2e^x/((1-x)^2))

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

t a y l or sin (3 x), \frac{π}{3}

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