derivative of sqrt((x^4-x+1/(x^4+x+1)))

Lösung

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

\frac{3 x ^{4} - 1}{x ^{4} - x + 1 ( x ^{4} + x + 1 ) ^{\frac{3}{2}}}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

\frac{d}{d x} (\frac{x ^{4} - x + 1}{x ^{4} + x + 1}) = \frac{6 x ^{4} - 2}{( x ^{4} + x + 1 ) ^{2}}

= \frac{1}{2 \frac{x ^{4} - x + 1}{x ^{4} + x + 1}} \cdot \frac{6 x ^{4} - 2}{( x ^{4} + x + 1 ) ^{2}}

Vereinfache

\frac{1}{2 \frac{x ^{4} - x + 1}{x ^{4} + x + 1}} \cdot \frac{6 x ^{4} - 2}{( x ^{4} + x + 1 ) ^{2}} : \frac{3 x ^{4} - 1}{x ^{4} - x + 1 ( x ^{4} + x + 1 ) ^{\frac{3}{2}}}

= \frac{3 x ^{4} - 1}{x ^{4} - x + 1 ( x ^{4} + x + 1 ) ^{\frac{3}{2}}}

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