derivative of (6x/(2+x^2))

Lösung

\frac{d}{d x} (\frac{6 x}{2 + x ^{2}})

\frac{6 ( 2 - x ^{2} )}{( 2 + x ^{2} ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{6 x}{2 + x ^{2}})

Entferne die Konstante:

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

= 6 \frac{d}{d x} (\frac{x}{2 + x ^{2}})

Wende die Quotientenregel an:

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

= 6 \cdot \frac{\frac{d x}{d x} ( 2 + x ^{2} ) - \frac{d}{d x} ( 2 + x ^{2} ) x}{( 2 + x ^{2} ) ^{2}}

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

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

= 6 \cdot \frac{1 \cdot ( 2 + x ^{2} ) - 2 xx}{( 2 + x ^{2} ) ^{2}}

Vereinfache

6 \cdot \frac{1 \cdot ( 2 + x ^{2} ) - 2 xx}{( 2 + x ^{2} ) ^{2}} : \frac{6 ( 2 - x ^{2} )}{( 2 + x ^{2} ) ^{2}}

= \frac{6 ( 2 - x ^{2} )}{( 2 + x ^{2} ) ^{2}}

x y^{^{'}} = 2 y + x^{3}

\frac{\partial}{\partial x} (k)

\int \frac{1}{a x ^{2} + b x + c} d x

y^{^{'}} - (\frac{2 x}{1 + x ^{2}}) y = 0, y (0) = 4

d er i v a t i v e (t^{2} - t) (t^{2} + t)