derivative of (x+7/(x^3+x-6))

Lösung

\frac{d}{d x} (\frac{x + 7}{x ^{3} + x - 6})

\frac{- 2 x ^{3} - 21 x ^{2} - 13}{( x ^{3} + x - 6 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x + 7}{x ^{3} + x - 6})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x + 7 ) ( x ^{3} + x - 6 ) - \frac{d}{d x} ( x ^{3} + x - 6 ) ( x + 7 )}{( x ^{3} + x - 6 ) ^{2}}

\frac{d}{d x} (x + 7) = 1

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

= \frac{1 \cdot ( x ^{3} + x - 6 ) - ( 3 x ^{2} + 1 ) ( x + 7 )}{( x ^{3} + x - 6 ) ^{2}}

Vereinfache

\frac{1 \cdot ( x ^{3} + x - 6 ) - ( 3 x ^{2} + 1 ) ( x + 7 )}{( x ^{3} + x - 6 ) ^{2}} : \frac{- 2 x ^{3} - 21 x ^{2} - 13}{( x ^{3} + x - 6 ) ^{2}}

= \frac{- 2 x ^{3} - 21 x ^{2} - 13}{( x ^{3} + x - 6 ) ^{2}}

x \to 1 lim (- x^{2} + 1)

x \cdot \frac{d y}{d x} + 7 y = \frac{1}{x}

\int e^{5 x} sin (5 x) d x

a re a y = cos (x), x = \frac{π}{4}, x = \frac{π}{3}

\frac{d y}{d x} + y = 0