Discussion
Suppose \(f(x, y)\) is of class \(c^{2}\), then under the coordinate transformation \(x = r\cos\theta\), \(y = r\sin\theta\), the expression \(\dfrac{\partial^{2} f}{\partial x^{2}} + \dfrac{\partial^{2} f}{\partial y^{2}}\) transforms to :
Answer: D
The Laplacian becomes \(f_{rr} + \tfrac{1}{r}f_{r} + \tfrac{1}{r^{2}}f_{\theta\theta}\).
With \(x = r\cos\theta\), \(y = r\sin\theta\), the chain rule gives \(f_{r} = f_{x}\cos\theta + f_{y}\sin\theta\) and \(f_{\theta} = -rf_{x}\sin\theta + rf_{y}\cos\theta\). Differentiating again and combining, the mixed terms cancel and
\[ \begin{aligned} f_{xx} + f_{yy} &= f_{rr} + \frac{1}{r}f_{r} \\ &\quad + \frac{1}{r^{2}}f_{\theta\theta} \end{aligned} \]
A check: \(f = r^{2} = x^{2} + y^{2}\) has Laplacian 4, and the formula gives \(2 + \tfrac{1}{r}(2r) + 0 = 4\). Options A and B drop the first-order term \(\tfrac{1}{r}f_{r}\) (B gives 2 for this check), and C has it without the \(1/r\), which is not even dimensionally consistent.
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