derivative y=(x+5)/(x^3+x-9)

Lösung

ableitung von

y = \frac{x + 5}{x ^{3} + x - 9}

\frac{- 2 x ^{3} - 15 x ^{2} - 14}{( x ^{3} + x - 9 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (x + 5) = 1

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

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

Vereinfache

\frac{1 \cdot ( x ^{3} + x - 9 ) - ( 3 x ^{2} + 1 ) ( x + 5 )}{( x ^{3} + x - 9 ) ^{2}} : \frac{- 2 x ^{3} - 15 x ^{2} - 14}{( x ^{3} + x - 9 ) ^{2}}

= \frac{- 2 x ^{3} - 15 x ^{2} - 14}{( x ^{3} + x - 9 ) ^{2}}

d er i v a t i v e f (x) = \frac{6 x + x ^{2}}{( 3 + x ) ^{2}}

\int \frac{3 - 18 x}{1 - 9 x ^{2}} d x

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

\frac{d}{d t} (\frac{1}{t ^{2} + 1})

t an g e n t f (x) = x + 1 - ln (x), at x = 0.5