Find the derivative \( y=x^{2} \ln \left(x^{3}-2\right) \) Type answer \( y^{\prime}= \) ((numerator)/(denominator)+numerator)) A

To find the derivative of \( y = x^{2} \ln(x^{3} - 2) \), we can use the product rule of differentiation. Let \( u = x^{2} \) and \( v = \ln(x^{3} - 2) \). The product rule states that \( (uv)' = u'v + uv' \). 1. Differentiate \( u \): \( u' = 2x \). 2. Differentiate \( v \): Using the chain rule, \( v' = \frac{1}{x^{3} - 2} \cdot (3x^{2}) = \frac{3x^{2}}{x^{3} - 2} \). Now, applying the product rule: \[ y' = u'v + uv' = (2x) \ln(x^{3} - 2) + x^{2} \left(\frac{3x^{2}}{x^{3} - 2}\right) \] Combining these gives: \[ y' = 2x \ln(x^{3} - 2) + \frac{3x^{4}}{x^{3} - 2} \] So the final expression would be: \[ y' = \frac{(2x \ln(x^{3} - 2))(x^{3} - 2) + 3x^{4}}{x^{3} - 2} \] Therefore, the answer can be expressed as: \[ y' = \left( (2x \ln(x^{3} - 2))(x^{3} - 2) + 3x^{4} \right) / (x^{3} - 2) \]