derivative of \sqrt[3]{6+2x+x^3}

Let's perform the differentiation:

\frac{d}{d x} (\sqrt[3]{6 + 2 x + x^{3}})

Apply the differentiation rule:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{n - 1}

with f =

6 + 2 x + x^{3}

ⓘ The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (\sqrt[3]{6 + 2 x + x^{3}})

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

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

ⓘ Multiplying a fraction by a term multiplies its numerator:

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

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

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

■The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

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

\frac{\frac{d}{d x} (6 + 2 x + x^{3})}{3 \cdot \sqrt[3]{{(6 + 2 x + x^{3})}^{2}}}

\frac{\frac{d}{d x} (6 + 2 x + x^{3})}{3 \cdot \sqrt[3]{{(6 + 2 x + x^{3})}^{2}}}

ⓘ The derivative of a sum is the sum of the derivatives:

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

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

\frac{0 + \frac{d}{d x} (2 x) + \frac{d}{d x} (x^{3})}{3 \cdot \sqrt[3]{{(6 + 2 x + x^{3})}^{2}}}

ⓘ Adding 0 changes nothing:

0 + \frac{d}{d x} (2 x)

\frac{d}{d x} (2 x)

■A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (2 x)

2

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (x^{3})

3 x^{2}

\frac{2 + 3 x^{2}}{3 \cdot \sqrt[3]{{(6 + 2 x + x^{3})}^{2}}}

▶Derivative:

\frac{3 x^{2} + 2}{3 \cdot \sqrt[3]{{(x^{3} + 2 x + 6)}^{2}}}