derivative of 1/(\sqrt[3]{x^2+x+3})

Lösung

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

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

Schritte zur Lösung

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

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

\frac{d}{d y} (x - y^{3} + y^{2} sin (x))

d er i v a t i v e x^{3} + 5

y \to - 3 lim ((5 - y)^{\frac{4}{3}})

f (x) = \frac{x ^{2} + 1}{x ^{2} + 4}

\frac{d}{d x} (1 + 9 x)