derivative of (x+6/(x^3+x-8))

Lösung

\frac{d}{d x} (\frac{x + 6}{x ^{3} + x - 8})

\frac{- 2 x ^{3} - 18 x ^{2} - 14}{( x ^{3} + x - 8 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x + 6}{x ^{3} + x - 8})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x + 6 ) ( x ^{3} + x - 8 ) - \frac{d}{d x} ( x ^{3} + x - 8 ) ( x + 6 )}{( x ^{3} + x - 8 ) ^{2}}

\frac{d}{d x} (x + 6) = 1

\frac{d}{d x} (x^{3} + x - 8) = 3 x^{2} + 1

= \frac{1 \cdot ( x ^{3} + x - 8 ) - ( 3 x ^{2} + 1 ) ( x + 6 )}{( x ^{3} + x - 8 ) ^{2}}

Vereinfache

\frac{1 \cdot ( x ^{3} + x - 8 ) - ( 3 x ^{2} + 1 ) ( x + 6 )}{( x ^{3} + x - 8 ) ^{2}} : \frac{- 2 x ^{3} - 18 x ^{2} - 14}{( x ^{3} + x - 8 ) ^{2}}

= \frac{- 2 x ^{3} - 18 x ^{2} - 14}{( x ^{3} + x - 8 ) ^{2}}

d er i v a t i v e x^{2} - 2 x - 1

in v erse l a pl a ce \frac{( s + 4 )}{s ^{2} + 4 s + 29}

\int 1

d er i v a t i v e 4 sec^{2} (x) - 6 sec^{4} (x)

y^{^{''}} + 1 y = x e^{5 x}