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

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Quotientenregel an:

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

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

\frac{d x}{d x} = 1

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

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

Vereinfache

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

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

x \to 0 lim (\frac{- x ^{2}}{x ^{2}})

d er i v a t i v e f (x) = \frac{sin ( π - x )}{π - x}

x \to 0 lim (5^{x + h} - 5^{x})

y^{^{'}} - tan (x) = 0

d er i v a t i v e x - y