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

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Multipliziere aus

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

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

\int \frac{cos ^{3} ( 2 x )}{sin ( 2 x )} d x

d er i v a t i v e \frac{π}{4}

\int sin^{3} (4 x) cos^{2} (4 x) d x

\frac{\partial}{\partial x} (4 - x^{2})

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