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

Lösung

ableitung von

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 )}

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

\frac{\partial}{\partial y} (e^{- x} sin (yz))

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

t an g e n t f (x) = 4 - 3 ln (x), at x = 1

d er i v a t i v e y = 6 + 4 x csc (x)