Word Count : 720

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We are given: $$ x = e^r \cos\theta, \quad y = e^r \sin\theta $$ and $f = f(x, y)$ is continuously differentiable with continuous second partial derivatives. We are to show: $$ \frac{\partial^2 f}{\partial r^2} + \frac{\partial^2 f}{\partial \theta^2} = (x^2+y^2)\left(\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}\right) $$ --- ### Step 1: Compute first derivatives using the chain rule By the chain rule: $$ \frac{\partial f}{\partial r} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial r} $$ Compute derivatives of $x$ and $y$ with respect to $r$: $$ \frac{\partial x}{\partial r} = e^r \cos\theta = x $$ $$ \frac{\partial y}{\partial r} = e^r \sin\theta = y $$ Hence: $$ \frac{\partial f}{\partial r} = x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} \quad ...(1) $$ --- Similarly, for $\theta$: $$ \frac{\partial f}{\partial \theta} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial _______ ____ ____ ______ __________ ___ _____ _________ _____.
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