derivative f(x)=(x^2-4x+3)/(x+1)

Lösung

ableitung von

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

\frac{x ^{2} + 2 x - 7}{( x + 1 ) ^{2}}

Schritte zur Lösung

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

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

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

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

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

Vereinfache

\frac{( 2 x - 4 ) ( x + 1 ) - 1 \cdot ( x ^{2} - 4 x + 3 )}{( x + 1 ) ^{2}} : \frac{x ^{2} + 2 x - 7}{( x + 1 ) ^{2}}

= \frac{x ^{2} + 2 x - 7}{( x + 1 ) ^{2}}