1. **State the problem:** Find the partial derivative $\frac{\partial z}{\partial x}$ for the function $$z = x^4 \sin(x y^3).$$ 2. **Recall the formula:** To find $\frac{\partial z}{\partial x}$, treat $y$ as a constant and differentiate with respect to $x$. Use the product rule: $$\frac{d}{dx}[u v] = u' v + u v'$$ where $$u = x^4$$ and $$v = \sin(x y^3).$$ 3. **Differentiate each part:** - $$u' = \frac{d}{dx} x^4 = 4x^3$$ - $$v' = \frac{d}{dx} \sin(x y^3) = \cos(x y^3) \cdot \frac{d}{dx}(x y^3) = \cos(x y^3) \cdot y^3$$ 4. **Apply the product rule:** $$\frac{\partial z}{\partial x} = u' v + u v' = 4x^3 \sin(x y^3) + x^4 y^3 \cos(x y^3).$$ 5. **Interpretation:** The derivative consists of two terms: the derivative of $x^4$ times the sine function plus $x^4$ times the derivative of the sine function with respect to $x$. **Final answer:** $$\boxed{4x^3 \sin(x y^3) + x^4 y^3 \cos(x y^3)}$$ which corresponds to option B.