derivative of (x^2+4x+5x^3/(x+1))

Lösung

\frac{d}{d x} (\frac{x ^{2} + 4 x + 5 x ^{3}}{x + 1})

\frac{10 x ^{3} + 16 x ^{2} + 2 x + 4}{( x + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{2} + 4 x + 5 x ^{3}}{x + 1})

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (x^{2} + 4 x + 5 x^{3}) = 2 x + 4 + 15 x^{2}

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

= \frac{( 2 x + 4 + 15 x ^{2} ) ( x + 1 ) - 1 \cdot ( x ^{2} + 4 x + 5 x ^{3} )}{( x + 1 ) ^{2}}

Vereinfache

\frac{( 2 x + 4 + 15 x ^{2} ) ( x + 1 ) - 1 \cdot ( x ^{2} + 4 x + 5 x ^{3} )}{( x + 1 ) ^{2}} : \frac{10 x ^{3} + 16 x ^{2} + 2 x + 4}{( x + 1 ) ^{2}}

= \frac{10 x ^{3} + 16 x ^{2} + 2 x + 4}{( x + 1 ) ^{2}}

\int_{1}^{2} 2^{- x} d x

\int_{0}^{1} \frac{25}{4 y - 1} d y

x \to 0 + lim (x ln (x + x^{2}))

\frac{d P}{d x} = P - P^{2}

\frac{d}{d x} (3 x - 6)