1. The problem states the operator $$H(r) = \frac{\partial}{\partial r} \left( \frac{1}{r} \frac{\partial}{\partial r} (r h) \right)$$ where $$r = \sqrt{x^2 + y^2}$$. 2. This operator is a radial part of the Laplacian in cylindrical or polar coordinates applied to the function $$h(r)$$. 3. To understand and simplify $$H(r)$$, we first compute the inner derivative: $$\frac{\partial}{\partial r} (r h) = h + r \frac{\partial h}{\partial r}$$ 4. Substitute back: $$H(r) = \frac{\partial}{\partial r} \left( \frac{1}{r} (h + r \frac{\partial h}{\partial r}) \right) = \frac{\partial}{\partial r} \left( \frac{h}{r} + \frac{\partial h}{\partial r} \right)$$ 5. Differentiate term by term: $$H(r) = \frac{\partial}{\partial r} \left( \frac{h}{r} \right) + \frac{\partial^2 h}{\partial r^2} = \frac{r \frac{\partial h}{\partial r} - h}{r^2} + \frac{\partial^2 h}{\partial r^2}$$ 6. Therefore, the full expression is: $$H(r) = \frac{\partial^2 h}{\partial r^2} + \frac{1}{r} \frac{\partial h}{\partial r} - \frac{h}{r^2}$$ 7. This formula is useful in problems involving radial symmetry, such as heat conduction or wave equations in cylindrical coordinates. Since the user provided values a. 25, b. 40, c. 16, and some other numbers without a clear question, we focus only on the first problem as per instructions. Final answer: $$H(r) = \frac{\partial^2 h}{\partial r^2} + \frac{1}{r} \frac{\partial h}{\partial r} - \frac{h}{r^2}$$