Find the derivative of \(\sqrt[3]{(3x^{3} + 1}\) with respect to x.

A.

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

B.

\(\frac{3x^{2}}{\sqrt[3]{(3x^{3} + 1)^{2}}}\)

C.

\(\frac{3x}{\sqrt[3]{3x^{2} + 1}}\)

D.

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

Correct answer is B

\(y = \sqrt[3]{3x^{3} + 1}  = (3x^{3} + 1)^{\frac{1}{3}}\)

Let u = \(3x^{3} + 1\); y = \(u^{\frac{1}{3}}\)

\(\frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)

\(\frac{\mathrm d y}{\mathrm d u} = \frac{1}{3}u^{\frac{-2}{3}}\)

\(\frac{\mathrm d u}{\mathrm d x} = 9x^{2}\)

\(\frac{\mathrm d y}{\mathrm d x} = (\frac{1}{3}(3x^{3} + 1)^{\frac{-2}{3}})(9x^{2})\)

= \(\frac{3x^{2}}{\sqrt[3]{(3x^{3} + 1)^{2}}}\)