设 r = sqrt(x^2 + y^2), u = f(r), 其中 f 具有二阶连续导函数, 则 (partial^2 u)/(partial x^2) + (partial^2 u)/(partial y^2) = ( ).A. f''(r)B. f''(r)+ (1)/(r) f'(r)C. f''(r)- (1)/(r) f'(r)D. r^2 f''(r)

步骤 1：计算一阶偏导数
给定 $r = \sqrt{x^2 + y^2}$ 和 $u = f(r)$，首先计算 $u$ 对 $x$ 和 $y$ 的一阶偏导数。根据链式法则，我们有：
\[ \frac{\partial u}{\partial x} = f'(r) \cdot \frac{\partial r}{\partial x} = f'(r) \cdot \frac{x}{r} \]
\[ \frac{\partial u}{\partial y} = f'(r) \cdot \frac{\partial r}{\partial y} = f'(r) \cdot \frac{y}{r} \]

步骤 2：计算二阶偏导数
接下来，计算 $u$ 对 $x$ 和 $y$ 的二阶偏导数。再次应用链式法则，我们得到：
\[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left( f'(r) \cdot \frac{x}{r} \right) = f''(r) \cdot \frac{x^2}{r^2} + f'(r) \cdot \frac{y^2}{r^3} \]
\[ \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y} \left( f'(r) \cdot \frac{y}{r} \right) = f''(r) \cdot \frac{y^2}{r^2} + f'(r) \cdot \frac{x^2}{r^3} \]

步骤 3：求和
最后，将二阶偏导数相加，得到：
\[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = f''(r) \cdot \frac{x^2}{r^2} + f'(r) \cdot \frac{y^2}{r^3} + f''(r) \cdot \frac{y^2}{r^2} + f'(r) \cdot \frac{x^2}{r^3} \]
\[ = f''(r) \cdot \left( \frac{x^2}{r^2} + \frac{y^2}{r^2} \right) + f'(r) \cdot \left( \frac{y^2}{r^3} + \frac{x^2}{r^3} \right) \]
\[ = f''(r) + \frac{1}{r} f'(r) \]