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

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

\int (2^{x} + 3^{x})^{2} d x

\frac{d y}{d t} = 6 y + t e^{6 t}

\frac{d}{d x} (x^{4} - 3 x^{2} + 5 x + 2)

x \to \infty lim (- 3 + \frac{4}{x})

\frac{d}{d x} (sec (x))