d/(ds)(((s-3)(s-5))/((s+1)(s+2)))

Lösung

\frac{d}{d s} (\frac{( s - 3 ) ( s - 5 )}{( s + 1 ) ( s + 2 )})

\frac{11 s ^{2} - 26 s - 61}{( s + 1 ) ^{2} ( s + 2 ) ^{2}}

Schritte zur Lösung

\frac{d}{d s} (\frac{( s - 3 ) ( s - 5 )}{( s + 1 ) ( s + 2 )})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d s} ( ( s - 3 ) ( s - 5 ) ) ( s + 1 ) ( s + 2 ) - \frac{d}{d s} ( ( s + 1 ) ( s + 2 ) ) ( s - 3 ) ( s - 5 )}{( ( s + 1 ) ( s + 2 ) ) ^{2}}

\frac{d}{d s} ((s - 3) (s - 5)) = 2 s - 8

\frac{d}{d s} ((s + 1) (s + 2)) = 2 s + 3

= \frac{( 2 s - 8 ) ( s + 1 ) ( s + 2 ) - ( 2 s + 3 ) ( s - 3 ) ( s - 5 )}{( ( s + 1 ) ( s + 2 ) ) ^{2}}

Vereinfache

\frac{( 2 s - 8 ) ( s + 1 ) ( s + 2 ) - ( 2 s + 3 ) ( s - 3 ) ( s - 5 )}{( ( s + 1 ) ( s + 2 ) ) ^{2}} : \frac{11 s ^{2} - 26 s - 61}{( s + 1 ) ^{2} ( s + 2 ) ^{2}}

= \frac{11 s ^{2} - 26 s - 61}{( s + 1 ) ^{2} ( s + 2 ) ^{2}}

\frac{d}{d t} (3 - 2 t)

d er i v a t i v e \frac{1 - x ^{2} + x ^{4}}{x ^{2} \cdot x}

x \to 0 lim (\frac{0}{0})

\int \frac{2 x ^{2} + x - 1}{2 x ^{3} + x ^{2} - 5 x + 2} d x

\int \frac{x - 1}{x ^{2} - 4} d x