Use the product rule to obtain the derivative of \( j(x)=3 x^{5}\left(3 x^{2}+x\right) \). \( j^{\prime}(x)=\square \)

To find the derivative of \( j(x) = 3x^{5}(3x^{2} + x) \) using the product rule, we start by identifying the two functions: let \( u = 3x^{5} \) and \( v = 3x^{2} + x \). The product rule states that \( (uv)' = u'v + uv' \). First, calculate \( u' = 15x^{4} \) and \( v' = 6x + 1 \). Now applying the product rule: \[ j'(x) = u'v + uv' = (15x^{4})(3x^{2} + x) + (3x^{5})(6x + 1) \] Now, we can simplify this expression: \[ j'(x) = 15x^{4}(3x^{2}) + 15x^{4}(x) + 18x^{6} + 3x^{5} \] Combine the like terms to finish: \[ j'(x) = 45x^{6} + 15x^{5} + 18x^{6} + 3x^{5} = 63x^{6} + 18x^{5} \] Therefore, the derivative is \( j'(x) = 63x^{6} + 18x^{5} \).