derivative y=((x+3)(x^2-3x+9))/(x^3)

Lösung

ableitung von

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

- \frac{81}{x ^{4}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

\frac{3 x ^{2} x ^{3} - 3 x ^{2} ( x + 3 ) ( x ^{2} - 3 x + 9 )}{( x ^{3} ) ^{2}} : - \frac{81}{x ^{4}}

= - \frac{81}{x ^{4}}

d er i v a t i v e ln (x + 6)

y^{^{'}} = 9 y^{2} + x y^{2}

\frac{d}{d x} (\frac{2 x}{x - 1})

d er i v a t i v e y = e^{4} x^{2} + 18

d er i v a t i v e f (x) = (\frac{8 x + 7}{9 x ^{2} + 6})