derivative of (sin(2x)/(1+cos(x)))

Lösung

\frac{d}{d x} (\frac{sin ( 2 x )}{1 + cos ( x )})

\frac{2 cos ( 2 x ) ( 1 + cos ( x ) ) + sin ( x ) sin ( 2 x )}{( 1 + cos ( x ) ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{sin ( 2 x )}{1 + cos ( x )})

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (sin (2 x)) = cos (2 x) \cdot 2

\frac{d}{d x} (1 + cos (x)) = - sin (x)

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

Vereinfache

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

\frac{\partial}{\partial x} (cos (x y) e^{x^{2}})

\int \frac{8 + 6 x}{1 + x ^{2}} d x

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