1. **Problem Statement:** Find the second partial derivatives $f_{12}$ and $f_{22}$ of the function $$f(x,y) = e^x \sin x + \ln(xy)$$ 2. **Formulas and Rules:** - The mixed partial derivative $f_{12}$ means first differentiate with respect to $x$, then with respect to $y$. - The second partial derivative $f_{22}$ means differentiate twice with respect to $y$. - Recall that $\frac{\partial}{\partial x} e^x \sin x = e^x \sin x + e^x \cos x$ by product rule. - For $\ln(xy)$, use $\frac{\partial}{\partial x} \ln(xy) = \frac{1}{xy} \cdot y = \frac{1}{x}$ and similarly for $y$. 3. **Step-by-step solution:** **Step 1: Compute $f_x$ (partial derivative with respect to $x$):** $$f_x = \frac{\partial}{\partial x} \left(e^x \sin x + \ln(xy)\right) = \frac{\partial}{\partial x} (e^x \sin x) + \frac{\partial}{\partial x} \ln(xy)$$ Using product rule: $$\frac{\partial}{\partial x} (e^x \sin x) = e^x \sin x + e^x \cos x$$ For the logarithm term: $$\frac{\partial}{\partial x} \ln(xy) = \frac{1}{xy} \cdot y = \frac{1}{x}$$ So, $$f_x = e^x \sin x + e^x \cos x + \frac{1}{x}$$ **Step 2: Compute $f_{12} = \frac{\partial}{\partial y} f_x$:** Since $e^x \sin x$ and $e^x \cos x$ do not depend on $y$, their derivatives w.r.t. $y$ are zero. For $\frac{1}{x}$, it does not depend on $y$, so derivative is zero. But we must check carefully: the term $\frac{1}{x}$ came from $\frac{\partial}{\partial x} \ln(xy)$, but $\ln(xy)$ depends on $y$. Actually, $f_x$ is: $$f_x = e^x \sin x + e^x \cos x + \frac{1}{x}$$ No $y$ terms remain explicitly, so $$f_{12} = \frac{\partial}{\partial y} f_x = 0$$ **Step 3: Compute $f_y$ (partial derivative with respect to $y$):** $$f_y = \frac{\partial}{\partial y} \left(e^x \sin x + \ln(xy)\right) = 0 + \frac{\partial}{\partial y} \ln(xy)$$ Since $e^x \sin x$ does not depend on $y$, derivative is zero. For $\ln(xy)$: $$\frac{\partial}{\partial y} \ln(xy) = \frac{1}{xy} \cdot x = \frac{1}{y}$$ So, $$f_y = \frac{1}{y}$$ **Step 4: Compute $f_{22} = \frac{\partial}{\partial y} f_y$: $$f_{22} = \frac{\partial}{\partial y} \left(\frac{1}{y}\right) = -\frac{1}{y^2}$$ 4. **Final answers:** $$f_{12} = 0$$ $$f_{22} = -\frac{1}{y^2}$$