derivative of (2x^3-4x^2+3/x)

Lösung

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

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

Schritte zur Lösung

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

\frac{d}{d x} (2 x^{3} - 4 x^{2} + 3) = 6 x^{2} - 8 x

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

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

Vereinfache

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

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

\frac{\partial}{\partial x} (- 9 e^{- 5 x y})

\int_{0}^{\frac{π}{6}} cos^{2} (3 x) d x

\int cot^{3} (8 x) d x

\int \frac{3}{2} sin (x) d x

t an g e n t f (x) = 2 x^{2} - 5 x + 3, at a = 1