derivative (x^2-2x+3)/(x+1)

Lösung

ableitung von

\frac{x ^{2} - 2 x + 3}{x + 1}

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

Schritte zur Lösung

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

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

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

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

Vereinfache

\frac{( 2 x - 2 ) ( x + 1 ) - 1 \cdot ( x ^{2} - 2 x + 3 )}{( x + 1 ) ^{2}} : \frac{x ^{2} + 2 x - 5}{( x + 1 ) ^{2}}

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

\int \frac{1}{7} d x

\int sin (2 x) ln (sin (2 x)) d x

in v erse l a pl a ce \frac{10}{s ( s ^{2} + 2 s + 10 )}

\frac{\partial}{\partial y} (x y + y^{2} + y)

t an g e n t f (x) = \frac{2}{x}, at (\frac{1}{9})