derivative of (x^3+2x^2-1/(x+5))

Lösung

\frac{d}{d x} (\frac{x ^{3} + 2 x ^{2} - 1}{x + 5})

\frac{2 x ^{3} + 17 x ^{2} + 20 x + 1}{( x + 5 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{3} + 2 x ^{2} - 1}{x + 5})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x ^{3} + 2 x ^{2} - 1 ) ( x + 5 ) - \frac{d}{d x} ( x + 5 ) ( x ^{3} + 2 x ^{2} - 1 )}{( x + 5 ) ^{2}}

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

\frac{d}{d x} (x + 5) = 1

= \frac{( 3 x ^{2} + 4 x ) ( x + 5 ) - 1 \cdot ( x ^{3} + 2 x ^{2} - 1 )}{( x + 5 ) ^{2}}

Vereinfache

\frac{( 3 x ^{2} + 4 x ) ( x + 5 ) - 1 \cdot ( x ^{3} + 2 x ^{2} - 1 )}{( x + 5 ) ^{2}} : \frac{2 x ^{3} + 17 x ^{2} + 20 x + 1}{( x + 5 ) ^{2}}

= \frac{2 x ^{3} + 17 x ^{2} + 20 x + 1}{( x + 5 ) ^{2}}

\int \frac{1}{2} x^{2} \cdot \frac{1}{x} d x

x \to \infty lim ((\frac{x ^{p}}{e ^{x}})^{\frac{1}{n}})

x \cdot \frac{d y}{d x} + 6 y = x^{3} - x

\frac{d}{d t} (4)

(\frac{2 t}{t ^{2} + 1})^{^{'}}