derivative of (2x^2+5x+8/(2x^2))

Lösung

\frac{d}{d x} (\frac{2 x ^{2} + 5 x + 8}{2 x ^{2}})

\frac{- 5 x - 16}{2 x ^{3}}

Schritte zur Lösung

\frac{d}{d x} (\frac{2 x ^{2} + 5 x + 8}{2 x ^{2}})

Entferne die Konstante:

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

= \frac{1}{2} \frac{d}{d x} (\frac{2 x ^{2} + 5 x + 8}{x ^{2}})

Wende die Quotientenregel an:

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

= \frac{1}{2} \cdot \frac{\frac{d}{d x} ( 2 x ^{2} + 5 x + 8 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( 2 x ^{2} + 5 x + 8 )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (2 x^{2} + 5 x + 8) = 4 x + 5

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

= \frac{1}{2} \cdot \frac{( 4 x + 5 ) x ^{2} - 2 x ( 2 x ^{2} + 5 x + 8 )}{( x ^{2} ) ^{2}}

Vereinfache

\frac{1}{2} \cdot \frac{( 4 x + 5 ) x ^{2} - 2 x ( 2 x ^{2} + 5 x + 8 )}{( x ^{2} ) ^{2}} : \frac{- 5 x - 16}{2 x ^{3}}

= \frac{- 5 x - 16}{2 x ^{3}}