((x^2+1)/(x^2-1))^'

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

(x^{2} + 1)^{'} = 2 x

(x^{2} - 1)^{'} = 2 x

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

Vereinfache

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

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

x \to 2.99 lim (1 + x)

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