derivative y=ln((x^2-1)/(x^2+1))

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

d er i v a t i v e - 10 e^{- 5 x^{2}} x

\int 6 x ln (5 x) d x

a re a y = x^{4} - 4 x^{2} + 4, y = x^{2}

in v erse l a pl a ce \frac{4}{s ^{2} ( s + 1 )}

y^{^{'}} = \frac{x ^{3} - 2 y}{x}