derivative of 1/(\sqrt[3]{x^2+x+5})

Lösung

\frac{d}{d x} (\frac{1}{3 x ^{2} + x + 5})

- \frac{2 x + 1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}}

Schritte zur Lösung

\frac{d}{d x} (\frac{1}{3 x ^{2} + x + 5})

Wende Exponentenregel an:

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

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

Wende die Kettenregel an:

- \frac{1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}} \frac{d}{d x} (x^{2} + x + 5)

= - \frac{1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}} \frac{d}{d x} (x^{2} + x + 5)

\frac{d}{d x} (x^{2} + x + 5) = 2 x + 1

= - \frac{1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}} (2 x + 1)

Vereinfache

- \frac{1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}} (2 x + 1) : - \frac{2 x + 1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}}

= - \frac{2 x + 1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}}

\int_{0}^{2} e^{x} - e^{- x} d x

s l o p e y = 3 x^{2} + 2 xx a

a re a y = e^{- 8 x}, x \geq 0

\frac{d}{d x} (sin (7 x) + sin^{2} (7 x))

n = 1 \sum \infty \frac{1}{n ^{2} + 2 n + 2}