derivative of ln((x^3-2x/(4+x)))

Lösung

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

\frac{2 x ^{3} + 12 x ^{2} - 8}{( x ^{3} - 2 x ) ( x + 4 )}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

\frac{d}{d x} (\frac{x ^{3} - 2 x}{4 + x}) = \frac{2 x ^{3} + 12 x ^{2} - 8}{( 4 + x ) ^{2}}

= \frac{1}{\frac{x ^{3} - 2 x}{4 + x}} \cdot \frac{2 x ^{3} + 12 x ^{2} - 8}{( 4 + x ) ^{2}}

Vereinfache

\frac{1}{\frac{x ^{3} - 2 x}{4 + x}} \cdot \frac{2 x ^{3} + 12 x ^{2} - 8}{( 4 + x ) ^{2}} : \frac{2 x ^{3} + 12 x ^{2} - 8}{( x ^{3} - 2 x ) ( x + 4 )}

= \frac{2 x ^{3} + 12 x ^{2} - 8}{( x ^{3} - 2 x ) ( x + 4 )}

d er i v a t i v e f (x) = \frac{2 x + 1}{x + 2}

\frac{d}{d x} (\frac{x}{2})

11 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 9

d er i v a t i v e y = (\frac{2}{( 2 x - 6 ) \frac{3}{2}})

t an g e n t - sin (\frac{x}{2})