derivative of ((x^3-1/(x^3+1))^7)

Lösung

\frac{d}{d x} ((\frac{x ^{3} - 1}{x ^{3} + 1})^{7})

\frac{42 x ^{2} ( x ^{3} - 1 ) ^{6}}{( x ^{3} + 1 ) ^{8}}

Schritte zur Lösung

\frac{d}{d x} ((\frac{x ^{3} - 1}{x ^{3} + 1})^{7})

Wende die Kettenregel an:

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

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

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

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

Vereinfache

7 (\frac{x ^{3} - 1}{x ^{3} + 1})^{6} \frac{6 x ^{2}}{( x ^{3} + 1 ) ^{2}} : \frac{42 x ^{2} ( x ^{3} - 1 ) ^{6}}{( x ^{3} + 1 ) ^{8}}

= \frac{42 x ^{2} ( x ^{3} - 1 ) ^{6}}{( x ^{3} + 1 ) ^{8}}

t an g e n t arctan (x + y) = y^{2} + \frac{π}{4}, (1, 0)

\int (x^{3} + 5 x^{2} - 2) e^{2 x} d x

y^{^{'}} = x^{2} - y - 2

\int (x - 1) (x^{2} - 2 x + 5)^{6} d x

d er i v a t i v e f (x) = 7 - x