1. **State the problem:** We are given the function $$f(x,y) = \sin(xy) + x^2 \ln(y)$$ and need to find the mixed partial derivative $$f_{yx}(0, \frac{\pi}{2})$$, which means first differentiate with respect to $$y$$, then with respect to $$x$$, and evaluate at the point $$(0, \frac{\pi}{2})$$. 2. **Recall the rules:** - The partial derivative $$f_y$$ means differentiate $$f$$ treating $$x$$ as constant. - The mixed partial $$f_{yx}$$ means $$\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)$$. - Use product and chain rules as needed. 3. **Find $$f_y$$:** $$f_y = \frac{\partial}{\partial y} \left( \sin(xy) + x^2 \ln(y) \right) = \cos(xy) \cdot x + x^2 \cdot \frac{1}{y}$$ 4. **Find $$f_{yx}$$:** Differentiate $$f_y$$ with respect to $$x$$: $$f_{yx} = \frac{\partial}{\partial x} \left( x \cos(xy) + \frac{x^2}{y} \right)$$ Apply product rule to $$x \cos(xy)$$: $$\frac{\partial}{\partial x} (x \cos(xy)) = \cos(xy) + x \cdot (-\sin(xy)) \cdot y = \cos(xy) - xy \sin(xy)$$ Differentiate $$\frac{x^2}{y}$$ with respect to $$x$$: $$\frac{\partial}{\partial x} \left( \frac{x^2}{y} \right) = \frac{2x}{y}$$ So, $$f_{yx} = \cos(xy) - xy \sin(xy) + \frac{2x}{y}$$ 5. **Evaluate at $$(0, \frac{\pi}{2})$$:** - $$\cos(0 \cdot \frac{\pi}{2}) = \cos(0) = 1$$ - $$0 \cdot \frac{\pi}{2} \cdot \sin(0) = 0$$ - $$\frac{2 \cdot 0}{\frac{\pi}{2}} = 0$$ Therefore, $$f_{yx}(0, \frac{\pi}{2}) = 1 + 0 + 0 = 1$$ **Final answer:** $$\boxed{1}$$