derivative (x+4)/(x^3+x-5)

Lösung

ableitung von

\frac{x + 4}{x ^{3} + x - 5}

\frac{- 2 x ^{3} - 12 x ^{2} - 9}{( x ^{3} + x - 5 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x + 4}{x ^{3} + x - 5})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x + 4 ) ( x ^{3} + x - 5 ) - \frac{d}{d x} ( x ^{3} + x - 5 ) ( x + 4 )}{( x ^{3} + x - 5 ) ^{2}}

\frac{d}{d x} (x + 4) = 1

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

= \frac{1 \cdot ( x ^{3} + x - 5 ) - ( 3 x ^{2} + 1 ) ( x + 4 )}{( x ^{3} + x - 5 ) ^{2}}

Vereinfache

\frac{1 \cdot ( x ^{3} + x - 5 ) - ( 3 x ^{2} + 1 ) ( x + 4 )}{( x ^{3} + x - 5 ) ^{2}} : \frac{- 2 x ^{3} - 12 x ^{2} - 9}{( x ^{3} + x - 5 ) ^{2}}

= \frac{- 2 x ^{3} - 12 x ^{2} - 9}{( x ^{3} + x - 5 ) ^{2}}

a re a y = 2 x, x = 5, y = \frac{2}{x}

\int \frac{x + 1}{x ^{3} + x ^{2} - 6 x} d x

t an g e n t y = x^{2}, at x = - 0.5

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

d er i v a t i v e y = (4 x^{6} + 7)^{5}