derivative of (2x+1/(1-x))

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

\int \frac{1}{4 x ^{2} - 1} d x

\frac{d}{d x} (x cos (y))

\int_{0}^{10} 2200 e^{0.024 x} - 1460 e^{0.018 x} d x

\int \frac{1}{T} d T

\frac{d}{d x} (sin^{2} (5 x - 3))