\( y = \sqrt[ 3 ] { x } ( 3 x ^ { 2 } - 2 - 4 x ^ { - 3 } ) \)

We start with \[ y = x^{\frac{1}{3}} \left( 3x^2 - 2 - 4x^{-3} \right). \] Our goal is to differentiate \( y \) with respect to \( x \). We will use the product rule, which states that if \[ y = u(x)v(x), \] then \[ y' = u'(x)v(x) + u(x)v'(x). \] Let \[ u(x) = x^{\frac{1}{3}} \quad \text{and} \quad v(x) = 3x^2 - 2 - 4x^{-3}. \] ### Step 1. Differentiate \( u(x) \) Using the power rule, the derivative of \( x^{\frac{1}{3}} \) is \[ u'(x) = \frac{1}{3} x^{\frac{1}{3}-1} = \frac{1}{3} x^{-\frac{2}{3}}. \] ### Step 2. Differentiate \( v(x) \) Differentiate each term in \( v(x) \): - For \( 3x^2 \): \[ \frac{d}{dx}(3x^2) = 6x. \] - The derivative of the constant \(-2\) is \(0\). - For \(-4x^{-3}\): \[ \frac{d}{dx}(-4x^{-3}) = -4(-3)x^{-4} = 12x^{-4}. \] Thus, \[ v'(x) = 6x + 12x^{-4}. \] ### Step 3. Apply the Product Rule Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the product rule: \[ y' = u'(x)v(x) + u(x)v'(x) = \frac{1}{3} x^{-\frac{2}{3}} \left( 3x^2 - 2 - 4x^{-3} \right) + x^{\frac{1}{3}} \left( 6x + 12x^{-4} \right). \] ### Step 4. Simplify Each Term First, simplify the expression \[ \frac{1}{3} x^{-\frac{2}{3}} \left( 3x^2 - 2 - 4x^{-3} \right) \] by distributing: - For the term with \( 3x^2 \): \[ \frac{1}{3} x^{-\frac{2}{3}} (3x^2) = x^{-\frac{2}{3}}x^2 = x^{\frac{4}{3}}. \] - For the term with \(-2\): \[ \frac{1}{3} x^{-\frac{2}{3}} (-2) = -\frac{2}{3} x^{-\frac{2}{3}}. \] - For the term with \(-4x^{-3}\): \[ \frac{1}{3} x^{-\frac{2}{3}}(-4x^{-3}) = -\frac{4}{3} x^{-\frac{2}{3}-3} = -\frac{4}{3} x^{-\frac{11}{3}}. \] Now, simplify the second part: \[ x^{\frac{1}{3}} \left( 6x + 12x^{-4} \right) \] by distributing: - For the term with \( 6x \): \[ x^{\frac{1}{3}} (6x) = 6x^{\frac{1}{3}+1} = 6x^{\frac{4}{3}}. \] - For the term with \( 12x^{-4} \): \[ x^{\frac{1}{3}} (12x^{-4}) = 12x^{\frac{1}{3}-4} = 12x^{-\frac{11}{3}}. \] ### Step 5. Combine Like Terms Now we combine the corresponding terms: \[ y' = \left( x^{\frac{4}{3}} + 6x^{\frac{4}{3}} \right) - \frac{2}{3} x^{-\frac{2}{3}} + \left( -\frac{4}{3}x^{-\frac{11}{3}} + 12x^{-\frac{11}{3}} \right). \] Combine the \( x^{\frac{4}{3}} \) terms: \[ x^{\frac{4}{3}} + 6x^{\frac{4}{3}} = 7x^{\frac{4}{3}}. \] Combine the \( x^{-\frac{11}{3}} \) terms: \[ -\frac{4}{3} x^{-\frac{11}{3}} + 12x^{-\frac{11}{3}} = \left( -\frac{4}{3} + 12 \right)x^{-\frac{11}{3}} = \frac{-4 + 36}{3}x^{-\frac{11}{3}} = \frac{32}{3}x^{-\frac{11}{3}}. \] Thus, the derivative is \[ y' = 7x^{\frac{4}{3}} - \frac{2}{3}x^{-\frac{2}{3}} + \frac{32}{3}x^{-\frac{11}{3}}. \]