derivative of (x^3-5x^2/(x^8-4))

Lösung

\frac{d}{d x} (\frac{x ^{3} - 5 x ^{2}}{x ^{8} - 4})

\frac{- 5 x ^{10} + 30 x ^{9} - 12 x ^{2} + 40 x}{( x ^{8} - 4 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{3} - 5 x ^{2}}{x ^{8} - 4})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x ^{3} - 5 x ^{2} ) ( x ^{8} - 4 ) - \frac{d}{d x} ( x ^{8} - 4 ) ( x ^{3} - 5 x ^{2} )}{( x ^{8} - 4 ) ^{2}}

\frac{d}{d x} (x^{3} - 5 x^{2}) = 3 x^{2} - 10 x

\frac{d}{d x} (x^{8} - 4) = 8 x^{7}

= \frac{( 3 x ^{2} - 10 x ) ( x ^{8} - 4 ) - 8 x ^{7} ( x ^{3} - 5 x ^{2} )}{( x ^{8} - 4 ) ^{2}}

Multipliziere aus

(3 x^{2} - 10 x) (x^{8} - 4) - 8 x^{7} (x^{3} - 5 x^{2}) : - 5 x^{10} + 30 x^{9} - 12 x^{2} + 40 x

= \frac{- 5 x ^{10} + 30 x ^{9} - 12 x ^{2} + 40 x}{( x ^{8} - 4 ) ^{2}}

\frac{d}{d x} (1 + sin^{2} (x))

d er i v a t i v e (2 t + 4) i + sin (4 t) j + 2 t^{3} k

x \to 0 + lim (x^{3 x})

d er i v a t i v e y = \frac{2}{x}

d er i v a t i v e y = \frac{13}{ln ( x )}