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

Lösung

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

\frac{e ^{x} ( 2 x ^{2} + 5 ) - 4 x e ^{x}}{( 2 x ^{2} + 5 ) ^{2}}

Schritte zur Lösung

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

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

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

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

= \frac{e ^{x} ( 2 x ^{2} + 5 ) - 4 x e ^{x}}{( 2 x ^{2} + 5 ) ^{2}}

\int_{- 1}^{1} 2 d x

b \to 0 lim (\frac{b}{x ^{2} + b ^{2}})

n = 3 \sum \infty \frac{4 ^{n} + 1}{n !}

d er i v a t i v e y = \frac{1}{( 1 + x ^{2} )}

s im pl i f y 3 x^{- 2} - 4 x