**Problem:** Find the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ for the function $f(x,y) = e^{xy} \sin(4y^2)$. 1. **Write down the function:** $$f(x,y) = e^{xy} \sin(4y^2)$$ 2. **Find $\frac{\partial f}{\partial x}$:** Since $f$ is a product of two functions, treat $\sin(4y^2)$ as a constant with respect to $x$. The derivative of $e^{xy}$ with respect to $x$ is found using chain rule: $$\frac{d}{dx} e^{xy} = e^{xy} \cdot y$$ Therefore, $$\frac{\partial f}{\partial x} = y e^{xy} \sin(4y^2)$$ 3. **Find $\frac{\partial f}{\partial y}$:** Use the product rule since both factors depend on $y$: $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left(e^{xy} \right) \sin(4y^2) + e^{xy} \frac{\partial}{\partial y} \left( \sin(4y^2) \right)$$ Derivative of the first part: $$\frac{\partial}{\partial y} e^{xy} = e^{xy} \cdot x$$ Derivative of the second part using chain rule: $$\frac{\partial}{\partial y} \sin(4y^2) = \cos(4y^2) \cdot 8y$$ So, $$\frac{\partial f}{\partial y} = x e^{xy} \sin(4y^2) + e^{xy} \cdot 8y \cos(4y^2)$$ **Final answers:** $$\boxed{\frac{\partial f}{\partial x} = y e^{xy} \sin(4y^2)}$$ $$\boxed{\frac{\partial f}{\partial y} = x e^{xy} \sin(4y^2) + 8y e^{xy} \cos(4y^2)}$$