derivative of (x^2+4x+3/(x^2+x))

Lösung

\frac{d}{d x} (\frac{x ^{2} + 4 x + 3}{x ^{2} + x})

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

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{2} + 4 x + 3}{x ^{2} + x})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x ^{2} + 4 x + 3 ) ( x ^{2} + x ) - \frac{d}{d x} ( x ^{2} + x ) ( x ^{2} + 4 x + 3 )}{( x ^{2} + x ) ^{2}}

\frac{d}{d x} (x^{2} + 4 x + 3) = 2 x + 4

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

= \frac{( 2 x + 4 ) ( x ^{2} + x ) - ( 2 x + 1 ) ( x ^{2} + 4 x + 3 )}{( x ^{2} + x ) ^{2}}

Vereinfache

\frac{( 2 x + 4 ) ( x ^{2} + x ) - ( 2 x + 1 ) ( x ^{2} + 4 x + 3 )}{( x ^{2} + x ) ^{2}} : - \frac{3}{x ^{2}}

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

\frac{d}{d x} (2 sec (3 x))

\frac{d}{d x} (\frac{1}{16} x^{4} + \frac{1}{2 x ^{2}})

x \to 3 lim (2 x + 3)

\frac{d}{d x} (r (x) \cdot cos (x))

n = 0 \sum \infty (- \frac{1}{2})^{n - 1}