1. Problem: Find the first partial derivatives of $$z = x^3 - 3x^2 y^4 + y^2$$ with respect to $$x$$ and $$y$$. 2. Formula: The partial derivative of $$z$$ with respect to $$x$$ is $$\frac{\partial z}{\partial x}$$, treating $$y$$ as a constant, and similarly for $$y$$. 3. Calculation: - $$\frac{\partial z}{\partial x} = 3x^2 - 6x y^4$$ (since $$y^4$$ is constant wrt $$x$$) - $$\frac{\partial z}{\partial y} = -12x^2 y^3 + 2y$$ (treating $$x$$ as constant) 4. Explanation: We differentiate each term with respect to the variable, holding the other variable constant. Final answer: $$\frac{\partial z}{\partial x} = 3x^2 - 6x y^4$$ $$\frac{\partial z}{\partial y} = -12x^2 y^3 + 2y$$