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

Lösung

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

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

Schritte zur Lösung

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

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

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

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

Multipliziere aus

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

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

d er i v a t i v e \frac{1}{2 r} + \frac{1}{9 r ^{(\frac{8}{9})}}

x \to 0 lim (\frac{1}{3 x ^{3}})

y^{^{''}} - 2 y^{^{'}} + y = t^{3} e^{t}

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

y^{^{''}} - 4 y^{^{'}} + 4 y = x e^{2 x}