derivative of (16x/((x^2+4)^2))

Lösung

\frac{d}{d x} (\frac{16 x}{( x ^{2} + 4 ) ^{2}})

\frac{16 ( - 3 x ^{2} + 4 )}{( x ^{2} + 4 ) ^{3}}

Schritte zur Lösung

\frac{d}{d x} (\frac{16 x}{( x ^{2} + 4 ) ^{2}})

Entferne die Konstante:

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

= 16 \frac{d}{d x} (\frac{x}{( x ^{2} + 4 ) ^{2}})

Wende die Quotientenregel an:

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

= 16 \cdot \frac{\frac{d x}{d x} ( x ^{2} + 4 ) ^{2} - \frac{d}{d x} ( ( x ^{2} + 4 ) ^{2} ) x}{( ( x ^{2} + 4 ) ^{2} ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} ((x^{2} + 4)^{2}) = 4 x (x^{2} + 4)

= 16 \cdot \frac{1 \cdot ( x ^{2} + 4 ) ^{2} - 4 x ( x ^{2} + 4 ) x}{( ( x ^{2} + 4 ) ^{2} ) ^{2}}

Vereinfache

16 \cdot \frac{1 \cdot ( x ^{2} + 4 ) ^{2} - 4 x ( x ^{2} + 4 ) x}{( ( x ^{2} + 4 ) ^{2} ) ^{2}} : \frac{16 ( - 3 x ^{2} + 4 )}{( x ^{2} + 4 ) ^{3}}

= \frac{16 ( - 3 x ^{2} + 4 )}{( x ^{2} + 4 ) ^{3}}

\frac{\partial}{\partial x} (3 arctan (\frac{y}{x}))

\int \frac{sec ( x ) tan ( x )}{sec ( x ) - 5} d x

a re a x^{2} - 7 x - 24, - x^{2} + 3 x + 4

\frac{\partial}{\partial y} (x - y)

\frac{d}{d x} (\frac{6}{3 x})