derivative of (x/(x^3+1)^6)

Lösung

\frac{d}{d x} ((\frac{x}{x ^{3} + 1})^{6})

\frac{6 x ^{5} ( - 2 x ^{3} + 1 )}{( x ^{3} + 1 ) ^{7}}

Schritte zur Lösung

\frac{d}{d x} ((\frac{x}{x ^{3} + 1})^{6})

Wende die Kettenregel an:

6 (\frac{x}{x ^{3} + 1})^{5} \frac{d}{d x} (\frac{x}{x ^{3} + 1})

= 6 (\frac{x}{x ^{3} + 1})^{5} \frac{d}{d x} (\frac{x}{x ^{3} + 1})

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

= 6 (\frac{x}{x ^{3} + 1})^{5} \frac{- 2 x ^{3} + 1}{( x ^{3} + 1 ) ^{2}}

Vereinfache

6 (\frac{x}{x ^{3} + 1})^{5} \frac{- 2 x ^{3} + 1}{( x ^{3} + 1 ) ^{2}} : \frac{6 x ^{5} ( - 2 x ^{3} + 1 )}{( x ^{3} + 1 ) ^{7}}

= \frac{6 x ^{5} ( - 2 x ^{3} + 1 )}{( x ^{3} + 1 ) ^{7}}

d er i v a t i v e 9 tan^{2} (6 x)

\frac{d}{d x} ((x^{2} + 15)^{2})

l a pl a ce t r an s f or m e^{2 t} cos (5 t)

d er i v a t i v e y = 3 x

a re a x^{2} + y + 4 = 0, y = - 8