1. Stating the problem: Given the vector function $$\mathbf{F} = (-3x^2 y - x^4)\mathbf{i} + (e^x y + 3 \sin x \cdot y)\mathbf{j} + (-2 x^2 \cos y)\mathbf{k},$$ we need to find the partial derivatives (i) $$\frac{\partial \mathbf{F}}{\partial x},$$ (ii) $$\frac{\partial \mathbf{F}}{\partial y},$$ and (iii) $$\frac{\partial^2 \mathbf{F}}{\partial x \partial y}.$$ 2. Compute $$\frac{\partial \mathbf{F}}{\partial x}$$ by differentiating each component with respect to $$x$$: - For the $$\mathbf{i}$$ component: $$-3x^2 y - x^4$$ $$\frac{\partial}{\partial x}(-3x^2 y - x^4) = -3 \cdot 2x y - 4x^3 = -6 x y - 4 x^3.$$ - For the $$\mathbf{j}$$ component: $$e^x y + 3 \sin x \cdot y$$ Use product rule noting $$y$$ is treated as constant: $$\frac{\partial}{\partial x}(e^x y) = e^x y,$$ $$\frac{\partial}{\partial x}(3 \sin x \cdot y) = 3 \cos x \cdot y.$$ So total: $$e^x y + 3 y \cos x.$$ - For the $$\mathbf{k}$$ component: $$-2 x^2 \cos y$$ Treat $$y$$ as constant: $$\frac{\partial}{\partial x}(-2 x^2 \cos y) = -4 x \cos y.$$ 3. Compute $$\frac{\partial \mathbf{F}}{\partial y}$$ by differentiating each component with respect to $$y$$: - For the $$\mathbf{i}$$ component: $$-3 x^2 y - x^4$$ $$\frac{\partial}{\partial y}(-3 x^2 y - x^4) = -3 x^2$$ (since $$x^4$$ is constant w.r.t. $$y$$). - For the $$\mathbf{j}$$ component: $$e^x y + 3 \sin x \cdot y$$ $$\frac{\partial}{\partial y}(e^x y + 3 \sin x \cdot y) = e^x + 3 \sin x.$$ - For the $$\mathbf{k}$$ component: $$-2 x^2 \cos y$$ $$\frac{\partial}{\partial y}(-2 x^2 \cos y) = -2 x^2 (-\sin y) = 2 x^2 \sin y.$$ 4. Compute $$\frac{\partial^2 \mathbf{F}}{\partial x \partial y}$$ by differentiating $$\frac{\partial \mathbf{F}}{\partial y}$$ with respect to $$x$$: - For $$\mathbf{i}$$ component: $$-3 x^2$$ $$\frac{\partial}{\partial x}(-3 x^2) = -6 x.$$ - For $$\mathbf{j}$$ component: $$e^x + 3 \sin x$$ $$\frac{\partial}{\partial x}(e^x + 3 \sin x) = e^x + 3 \cos x.$$ - For $$\mathbf{k}$$ component: $$2 x^2 \sin y$$ $$\frac{\partial}{\partial x}(2 x^2 \sin y) = 4 x \sin y$$ (since $$\sin y$$ is constant w.r.t. $$x$$). Final answers: $$\frac{\partial \mathbf{F}}{\partial x} = (-6 x y - 4 x^3)\mathbf{i} + (e^x y + 3 y \cos x)\mathbf{j} + (-4 x \cos y)\mathbf{k},$$ $$\frac{\partial \mathbf{F}}{\partial y} = (-3 x^2)\mathbf{i} + (e^x + 3 \sin x)\mathbf{j} + (2 x^2 \sin y)\mathbf{k},$$ $$\frac{\partial^2 \mathbf{F}}{\partial x \partial y} = (-6 x)\mathbf{i} + (e^x + 3 \cos x)\mathbf{j} + (4 x \sin y)\mathbf{k}.$$