derivative f(x)=(x^2+2x-4)/((x+1)^2)

Lösung

ableitung von

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

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

Schritte zur Lösung

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

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

Vereinfache

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

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

t an g e n t x^{3} - x^{2} + 1

\int x \cdot ln (\frac{1 - x}{1 + x}) d x

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

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

x \to \frac{π}{3} lim (tan (3 x))