1. **State the problem:** Find the first derivative of the function $z = x^3 \log x$ with respect to $x$. 2. **Recall the formula:** To differentiate a product of two functions, use the product rule: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ where $u(x) = x^3$ and $v(x) = \log x$. 3. **Differentiate each part:** - Derivative of $u(x) = x^3$ is $u'(x) = 3x^2$. - Derivative of $v(x) = \log x$ is $v'(x) = \frac{1}{x}$. 4. **Apply the product rule:** $$\frac{dz}{dx} = 3x^2 \cdot \log x + x^3 \cdot \frac{1}{x}$$ 5. **Simplify the expression:** $$\frac{dz}{dx} = 3x^2 \log x + x^2$$ 6. **Final answer:** $$\boxed{\frac{dz}{dx} = x^2 (3 \log x + 1)}$$