derivative f(x)=sqrt((3x+1)/(2x^2+1))

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

\frac{d}{d x} (\frac{3 x + 1}{2 x ^{2} + 1}) = \frac{- 6 x ^{2} - 4 x + 3}{( 2 x ^{2} + 1 ) ^{2}}

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

Vereinfache

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

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

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