derivative y=((x+4)(x^2-4x+16))/(x^3)

Lösung

ableitung von

y = \frac{( x + 4 ) ( x ^{2} - 4 x + 16 )}{x ^{3}}

- \frac{192}{x ^{4}}

Schritte zur Lösung

\frac{d}{d x} (\frac{( x + 4 ) ( x ^{2} - 4 x + 16 )}{x ^{3}})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( ( x + 4 ) ( x ^{2} - 4 x + 16 ) ) x ^{3} - \frac{d}{d x} ( x ^{3} ) ( x + 4 ) ( x ^{2} - 4 x + 16 )}{( x ^{3} ) ^{2}}

\frac{d}{d x} ((x + 4) (x^{2} - 4 x + 16)) = 3 x^{2}

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

= \frac{3 x ^{2} x ^{3} - 3 x ^{2} ( x + 4 ) ( x ^{2} - 4 x + 16 )}{( x ^{3} ) ^{2}}

Vereinfache

\frac{3 x ^{2} x ^{3} - 3 x ^{2} ( x + 4 ) ( x ^{2} - 4 x + 16 )}{( x ^{3} ) ^{2}} : - \frac{192}{x ^{4}}

= - \frac{192}{x ^{4}}

x \to 2 lim (\frac{2}{4 x + 3})

\int \frac{1}{( x - 5 ) ^{2}} - 1 d x

d er i v a t i v e \frac{1}{2 x}

\int t^{- 4} e^{t} d t

\frac{d}{d x} (7 x^{3} + 3 x - 12)