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

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Quotientenregel an:

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

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

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

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

= 2 \cdot \frac{( - \frac{s i n ( x )}{2 x} ) ( x + 1 ) - 1 \cdot cos ( x )}{( x + 1 ) ^{2}}

Vereinfache

2 \cdot \frac{( - \frac{s i n ( x )}{2 x} ) ( x + 1 ) - 1 \cdot cos ( x )}{( x + 1 ) ^{2}} : \frac{- sin ( x ) ( x + 1 ) - 2 x cos ( x )}{x ( x + 1 ) ^{2}}

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

\int \frac{( x - 2 )}{2} d x

\frac{d}{d x} (\frac{10 x}{x ^{2} - 6})

t an g e n t f (x) = x^{2} - 5, (- 3, 4)

\int (x - 3)^{2} \cdot e^{- x} d x

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