derivative of \sqrt[3]{x^3+6x+7}

Lösung

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

d er i v a t i v e f (x) = \frac{- 4}{x ^{3}}

\int \frac{x ^{3} + x ^{2} + x + 2}{x ^{4} + 3 x ^{2} + 2} d x

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

d er i v a t i v e \frac{5}{( 2 x ) ^{3}}

\int_{0}^{\frac{1}{4}} x arcsin (4 x) d x