derivative of ln((x^2-3x+2/(x^2+1)))

Lösung

\frac{d}{d x} (ln (\frac{x ^{2} - 3 x + 2}{x ^{2} + 1}))

\frac{3 x ^{2} - 2 x - 3}{( x ^{2} - 3 x + 2 ) ( x ^{2} + 1 )}

Schritte zur Lösung

\frac{d}{d x} (ln (\frac{x ^{2} - 3 x + 2}{x ^{2} + 1}))

Wende die Kettenregel an:

\frac{1}{\frac{x ^{2} - 3 x + 2}{x ^{2} + 1}} \frac{d}{d x} (\frac{x ^{2} - 3 x + 2}{x ^{2} + 1})

= \frac{1}{\frac{x ^{2} - 3 x + 2}{x ^{2} + 1}} \frac{d}{d x} (\frac{x ^{2} - 3 x + 2}{x ^{2} + 1})

\frac{d}{d x} (\frac{x ^{2} - 3 x + 2}{x ^{2} + 1}) = \frac{3 x ^{2} - 2 x - 3}{( x ^{2} + 1 ) ^{2}}

= \frac{1}{\frac{x ^{2} - 3 x + 2}{x ^{2} + 1}} \cdot \frac{3 x ^{2} - 2 x - 3}{( x ^{2} + 1 ) ^{2}}

Vereinfache

\frac{1}{\frac{x ^{2} - 3 x + 2}{x ^{2} + 1}} \cdot \frac{3 x ^{2} - 2 x - 3}{( x ^{2} + 1 ) ^{2}} : \frac{3 x ^{2} - 2 x - 3}{( x ^{2} - 3 x + 2 ) ( x ^{2} + 1 )}

= \frac{3 x ^{2} - 2 x - 3}{( x ^{2} - 3 x + 2 ) ( x ^{2} + 1 )}

s im pl i f y \frac{- 3 x ^{5} - x ^{4} - x ^{3} + 4 x ^{2}}{x ^{2}}

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