Find the derivative of \(\sqrt[3]{(3x^{3} + 1}\) with respect to x.
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Correct Answer: Option B
Explanation:
\(y = \sqrt[3]{3x^{3} + 1} = (3x^{3} + 1)^{\frac{1}{3}}\)
Let u = \(3x^{3} + 1\); y = \(u^{\frac{1}{3}}\)
\(\frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)
\(\frac{\mathrm d y}{\mathrm d u} = \frac{1}{3}u^{\frac{-2}{3}}\)
\(\frac{\mathrm d u}{\mathrm d x} = 9x^{2}\)
\(\frac{\mathrm d y}{\mathrm d x} = (\frac{1}{3}(3x^{3} + 1)^{\frac{-2}{3}})(9x^{2})\)
= \(\frac{3x^{2}}{\sqrt[3]{(3x^{3} + 1)^{2}}}\)
\(y = \sqrt[3]{3x^{3} + 1} = (3x^{3} + 1)^{\frac{1}{3}}\)
Let u = \(3x^{3} + 1\); y = \(u^{\frac{1}{3}}\)
\(\frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)
\(\frac{\mathrm d y}{\mathrm d u} = \frac{1}{3}u^{\frac{-2}{3}}\)
\(\frac{\mathrm d u}{\mathrm d x} = 9x^{2}\)
\(\frac{\mathrm d y}{\mathrm d x} = (\frac{1}{3}(3x^{3} + 1)^{\frac{-2}{3}})(9x^{2})\)
= \(\frac{3x^{2}}{\sqrt[3]{(3x^{3} + 1)^{2}}}\)