derivative of (4x^3-2x+1/(x^2))

Lösung

\frac{d}{d x} (\frac{4 x ^{3} - 2 x + 1}{x ^{2}})

\frac{2 ( 2 x ^{3} + x - 1 )}{x ^{3}}

Schritte zur Lösung

\frac{d}{d x} (\frac{4 x ^{3} - 2 x + 1}{x ^{2}})

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (4 x^{3} - 2 x + 1) = 12 x^{2} - 2

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

= \frac{( 12 x ^{2} - 2 ) x ^{2} - 2 x ( 4 x ^{3} - 2 x + 1 )}{( x ^{2} ) ^{2}}

Vereinfache

\frac{( 12 x ^{2} - 2 ) x ^{2} - 2 x ( 4 x ^{3} - 2 x + 1 )}{( x ^{2} ) ^{2}} : \frac{2 ( 2 x ^{3} + x - 1 )}{x ^{3}}

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

\int_{0}^{2 π} 3 sin (2 x) d x

x \to 0 lim (\frac{- ln ( 1 + 6 x )}{x})

\int \frac{x ^{3}}{1 + x ^{2} - x ^{4}} d x

\frac{d}{d y} (\frac{x y}{x ^{2} + y ^{2}})

d er i v a t i v e f (x) = \frac{2 - 5 x}{e ^{x}}