derivative of (e^x/(2x^2+9))

Lösung

\frac{d}{d x} (\frac{e ^{x}}{2 x ^{2} + 9})

\frac{e ^{x} ( 2 x ^{2} + 9 ) - 4 x e ^{x}}{( 2 x ^{2} + 9 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{e ^{x}}{2 x ^{2} + 9})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( e ^{x} ) ( 2 x ^{2} + 9 ) - \frac{d}{d x} ( 2 x ^{2} + 9 ) e ^{x}}{( 2 x ^{2} + 9 ) ^{2}}

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

\frac{d}{d x} (2 x^{2} + 9) = 4 x

= \frac{e ^{x} ( 2 x ^{2} + 9 ) - 4 x e ^{x}}{( 2 x ^{2} + 9 ) ^{2}}

x \cdot y \cdot y^{^{'}} = 1 - x^{2}

\frac{d}{d x} (\frac{x ^{3} - x}{x})

\frac{d}{d x} (\frac{4}{( x ^{3} + 9 ) ^{5}})

\frac{d x}{d t} + 2 t^{5} x^{7} + \frac{x}{t} = 0

n = 0 \sum \infty e^{- 6 n}