derivative y=(x+3)/(x^3+x-1)

Lösung

ableitung von

y = \frac{x + 3}{x ^{3} + x - 1}

\frac{- 2 x ^{3} - 9 x ^{2} - 4}{( x ^{3} + x - 1 ) ^{2}}

Schritte zur Lösung

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

\frac{d}{d x} (x + 3) = 1

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

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

Vereinfache

\frac{1 \cdot ( x ^{3} + x - 1 ) - ( 3 x ^{2} + 1 ) ( x + 3 )}{( x ^{3} + x - 1 ) ^{2}} : \frac{- 2 x ^{3} - 9 x ^{2} - 4}{( x ^{3} + x - 1 ) ^{2}}

= \frac{- 2 x ^{3} - 9 x ^{2} - 4}{( x ^{3} + x - 1 ) ^{2}}