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

Lösung

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

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

Schritte zur Lösung

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

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

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

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

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

\int - x^{- 4} d x

\int t^{4} d t

d er i v a t i v e y = x + 1

d er i v a t i v e f (x) = (x - 1) \cdot ln (x - 1)

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