derivative (x^2+x-2)/(x^3+6)

Lösung

ableitung von

\frac{x ^{2} + x - 2}{x ^{3} + 6}

\frac{- x ^{4} - 2 x ^{3} + 6 x ^{2} + 12 x + 6}{( x ^{3} + 6 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Multipliziere aus

(2 x + 1) (x^{3} + 6) - 3 x^{2} (x^{2} + x - 2) : - x^{4} - 2 x^{3} + 6 x^{2} + 12 x + 6

= \frac{- x ^{4} - 2 x ^{3} + 6 x ^{2} + 12 x + 6}{( x ^{3} + 6 ) ^{2}}

d er i v a t i v e y = \frac{x ^{2} - 4}{x - 3}

\frac{d}{d x} ((sec (x) + tan (x))^{- 3})

\int_{0}^{4} 5 x d x

x \to 0 + lim (ln (tan (x)))

\frac{d}{d x} (arcsec (x) - 7 x)