derivative of (e^x(x^2-4x+5)/((1-x)^3))

Lösung

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

\frac{e ^{x} ( - x ^{3} + 6 x ^{2} - 15 x + 16 )}{( 1 - x ) ^{4}}

Schritte zur Lösung

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

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

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

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

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

Vereinfache

\frac{( e ^{x} x ^{2} - 2 e ^{x} x + e ^{x} ) ( 1 - x ) ^{3} - ( - 3 ( 1 - x ) ^{2} ) e ^{x} ( x ^{2} - 4 x + 5 )}{( ( 1 - x ) ^{3} ) ^{2}} : \frac{e ^{x} ( - x ^{3} + 6 x ^{2} - 15 x + 16 )}{( 1 - x ) ^{4}}

= \frac{e ^{x} ( - x ^{3} + 6 x ^{2} - 15 x + 16 )}{( 1 - x ) ^{4}}

\int tan (x) d x

\int \frac{1}{x 25 - x ^{2}} d x

\int 5 x^{3} d x

\int 2 sec^{2} (x) tan^{3} (x) d x

f^{^{'}} (x) = 2 + 5 x^{2}