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

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

\frac{\partial}{\partial x} (x y + \frac{2}{x} + \frac{4}{y})

a re a e^{x}, e^{- 4 x}, x = ln (4)

\frac{d}{d x} (\frac{x - 2}{5 x ^{2} e ^{x}})

n = 1 \sum \infty \frac{1}{n ^{2} + 4 n + 3}

\int_{- 2}^{3} \frac{32}{x ^{4}} d x