derivative (4x^2-3)/(8x^3+1)

Lösung

ableitung von

\frac{4 x ^{2} - 3}{8 x ^{3} + 1}

\frac{- 32 x ^{4} + 72 x ^{2} + 8 x}{( 8 x ^{3} + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{4 x ^{2} - 3}{8 x ^{3} + 1})

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( 4 x ^{2} - 3 ) ( 8 x ^{3} + 1 ) - \frac{d}{d x} ( 8 x ^{3} + 1 ) ( 4 x ^{2} - 3 )}{( 8 x ^{3} + 1 ) ^{2}}

\frac{d}{d x} (4 x^{2} - 3) = 8 x

\frac{d}{d x} (8 x^{3} + 1) = 24 x^{2}

= \frac{8 x ( 8 x ^{3} + 1 ) - 24 x ^{2} ( 4 x ^{2} - 3 )}{( 8 x ^{3} + 1 ) ^{2}}

Multipliziere aus

8 x (8 x^{3} + 1) - 24 x^{2} (4 x^{2} - 3) : - 32 x^{4} + 72 x^{2} + 8 x

= \frac{- 32 x ^{4} + 72 x ^{2} + 8 x}{( 8 x ^{3} + 1 ) ^{2}}

\frac{d}{d x} (\frac{x cos ( π x ) + 1}{1 - x})

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

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\int (\frac{- 4}{x}) d x

\int \frac{1}{n} sin (n x) d x