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

Lösung

ableitung von

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

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

Schritte zur Lösung

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

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

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

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

Vereinfache

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

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

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