((sin(x))/(1+x))^'

Lösung

(\frac{sin ( x )}{1 + x})^{^{'}}

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

Schritte zur Lösung

(\frac{sin ( x )}{1 + x})^{'}

Wende die Quotientenregel an:

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

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

(sin (x))^{'} = cos (x)

(1 + x)^{'} = 1

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

Vereinfache

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

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (0) = - 2, y^{^{'}} (0) = - 4

(- x + 2) \cdot y^{^{'}} - y = e^{- 2 x}

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

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

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