derivative of (e^{2x-1}/((x^2+5x)))

Lösung

\frac{d}{d x} (\frac{e ^{2 x - 1}}{( x ^{2} + 5 x )})

\frac{2 e ^{2 x - 1} x ^{2} + 8 e ^{2 x - 1} x - 5 e ^{2 x - 1}}{( x ^{2} + 5 x ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{e ^{2 x - 1}}{( x ^{2} + 5 x )})

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (e^{2 x - 1}) = e^{2 x - 1} \cdot 2

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

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

Multipliziere aus

e^{2 x - 1} \cdot 2 (x^{2} + 5 x) - (2 x + 5) e^{2 x - 1} : 2 e^{2 x - 1} x^{2} + 8 e^{2 x - 1} x - 5 e^{2 x - 1}

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

d er i v a t i v e \frac{1}{4 y}

2 y d x + x d y = 0

\int \frac{( x ^{2} + 1 )}{x + 1} d x

t an g e n t f (x) = \frac{3}{4 x + 1}, at x = - 1

\int \frac{x - 4}{x ^{9}} d x