derivative of (2x^3+8/x)

Lösung

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

\frac{4 x ^{3} - 8}{x ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

\frac{d x}{d x} = 1

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

Vereinfache

\frac{6 x ^{2} x - 1 \cdot ( 2 x ^{3} + 8 )}{x ^{2}} : \frac{4 x ^{3} - 8}{x ^{2}}

= \frac{4 x ^{3} - 8}{x ^{2}}

\frac{d}{d x} (8 x cos (x^{2}))

x \to 8 lim ({f (7) g (- 7)})

\int \frac{4 x ^{2} + 8}{2 x ^{2} + 1} d x

\int \frac{1}{x 25 x ^{2} - 4} d x

\int \frac{e ^{y}}{1 - e ^{y}} d y