derivative f(x)=(6x^4-9x^2)/(3x^3)

Lösung

ableitung von

f (x) = \frac{6 x ^{4} - 9 x ^{2}}{3 x ^{3}}

\frac{2 x ^{2} + 3}{x ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{6 x ^{4} - 9 x ^{2}}{3 x ^{3}})

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{3} \frac{d}{d x} (\frac{6 x ^{4} - 9 x ^{2}}{x ^{3}})

Wende die Quotientenregel an:

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

= \frac{1}{3} \cdot \frac{\frac{d}{d x} ( 6 x ^{4} - 9 x ^{2} ) x ^{3} - \frac{d}{d x} ( x ^{3} ) ( 6 x ^{4} - 9 x ^{2} )}{( x ^{3} ) ^{2}}

\frac{d}{d x} (6 x^{4} - 9 x^{2}) = 24 x^{3} - 18 x

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

= \frac{1}{3} \cdot \frac{( 24 x ^{3} - 18 x ) x ^{3} - 3 x ^{2} ( 6 x ^{4} - 9 x ^{2} )}{( x ^{3} ) ^{2}}

Vereinfache

\frac{1}{3} \cdot \frac{( 24 x ^{3} - 18 x ) x ^{3} - 3 x ^{2} ( 6 x ^{4} - 9 x ^{2} )}{( x ^{3} ) ^{2}} : \frac{2 x ^{2} + 3}{x ^{2}}

= \frac{2 x ^{2} + 3}{x ^{2}}

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

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

\int \frac{5}{x 4 - ln ^{2} ( x )} d x

\frac{d}{d x} (- 12 x cos (x^{2}) + 8 x^{3} sin (x^{2}))

in v erse l a pl a ce \frac{1}{s ^{2} + 6 s + 13}