derivative (x^2+2x-7)/((x+1)^2)

Lösung

ableitung von

\frac{x ^{2} + 2 x - 7}{( x + 1 ) ^{2}}

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

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{2} + 2 x - 7}{( x + 1 ) ^{2}})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x ^{2} + 2 x - 7 ) ( x + 1 ) ^{2} - \frac{d}{d x} ( ( x + 1 ) ^{2} ) ( x ^{2} + 2 x - 7 )}{( ( x + 1 ) ^{2} ) ^{2}}

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

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

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

Vereinfache

\frac{( 2 x + 2 ) ( x + 1 ) ^{2} - 2 ( x + 1 ) ( x ^{2} + 2 x - 7 )}{( ( x + 1 ) ^{2} ) ^{2}} : \frac{16}{( x + 1 ) ^{3}}

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

\int (\frac{( ln ( x ) ) ^{2}}{x ^{2}}) d x

n = 0 \sum \infty 3 (\frac{1}{5})^{n}

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

\frac{\partial}{\partial x} (x y + z^{3} x)

\int e^{z}