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

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

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

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

\frac{\partial ^{2}}{\partial y ^{2}} (x y^{2} - 6 x^{2} - 3 y^{2})

\int_{0}^{1} e^{5 x} d x

y \to 2 lim (\frac{y ^{2} - 4}{y + 2})