This derivation was particular interesting because of the subtle mistakes one can make and still get the correct answer. This was one of the problems in my assigned homeworks, where I also made a mistake in one of the partial differentiation steps.

Given that the Laplace equation in Cartesian coordinates is given by

\[ \begin{align*} ∇^2 ϕ = \frac{\partial^2 ϕ}{\partial x^2} + \frac{\partial^2 ϕ}{\partial y^2} + \frac{\partial^2 ϕ}{\partial z^2} \\ \end{align*} \]

Let \(r\) be the hypotenuse of a triangle, \(x\) as the adjacent, and \(y\) as the opposite.

\[ \begin{align*} x = r \cos(\theta), \quad y = r \sin(\theta) \\ x^2 + y^2 = r^2 \\ \end{align*} \]

\[ \begin{align*} \frac{\partial} {\partial x} \left( r^2 \right) &= \frac{\partial }{\partial x} \left( x^2 + y^2 \right) \\ 2r \frac{\partial r}{\partial x} &= 2x \frac{\partial x}{\partial x} \\ \frac{\partial r}{\partial x} &= \frac{x}{r} = \frac{r \cos(\theta)}{r} = \cos(\theta) \\ \frac{\partial} {\partial y} \left( r^2 \right) &= \frac{\partial }{\partial y} \left( x^2 + y^2 \right) \\ 2r \frac{\partial r}{\partial y} &= 2y \frac{\partial y}{\partial y} \\ \frac{\partial r}{\partial y} &= \frac{y}{r} = \frac{r \sin(\theta)}{r} = \sin(\theta) \\ \frac{\partial z}{\partial x} &= \frac{\partial z}{\partial y} = 0 \\ \end{align*} \]

From trigonometry,

\[ \begin{align*} \tan( \theta ) &= \frac{y}{x} \\ \theta &= \tan ^{-1} \left( \frac{y}{x} \right) \\ \frac{\partial \theta}{\partial x} &= \frac{\partial}{\partial x} \left( \tan ^{-1} \left( \frac{y}{x} \right) \right) \\ &= \frac{1}{1 + \left( \frac{y}{x} \right)^2 } \left( - \frac{y}{x^2} \right) \\ &= \frac{- y}{x^2 + y^2} = \frac{-r \sin(\theta)}{r^2} = \frac{-\sin(\theta)}{r} \\ \theta &= \tan ^{-1} \left( \frac{y}{x} \right) \\ \frac{\partial \theta}{\partial y} &= \frac{\partial}{\partial y} \left( \tan ^{-1} \left( \frac{y}{x} \right) \right) \\ &= \frac{1}{1 + \left( \frac{y}{x} \right)^2 } \left( \frac{1}{x} \right) \\ &= \frac{x}{x^2 + y^2} = \frac{r \cos(\theta)}{r^2} = \frac{\cos(\theta)}{r} \\ \end{align*} \]

\[ \begin{align*} \frac{\partial ϕ}{\partial x} &= \frac{\partial ϕ}{ \partial r} \cdot \frac{\partial r}{\partial x} + \frac{\partial ϕ}{ \partial \theta} \cdot \frac{\partial \theta}{\partial x} + \frac{\partial ϕ}{ \partial z} \cdot \frac{\partial z}{\partial x} \\ & = \cos(\theta) \frac{\partial ϕ}{\partial r} + \frac{-\sin(\theta)}{r} \frac{\partial \phi}{\partial \theta} + 0 \\ \frac{\partial ϕ}{\partial y} &= \frac{\partial ϕ}{ \partial r} \cdot \frac{\partial r}{\partial y} + \frac{\partial ϕ}{ \partial \theta} \cdot \frac{\partial \theta}{\partial y} + \frac{\partial ϕ}{ \partial z} \cdot \frac{\partial z}{\partial y} \\ & = \sin(\theta) \frac{\partial ϕ}{\partial r} + \frac{\cos(\theta)}{r} \frac{\partial \phi}{\partial r} + 0 \\ \frac{\partial^2 ϕ}{\partial x^2} & = \cos(\theta) \frac{\partial^2 ϕ}{\partial r^2} \cdot \frac{\partial r}{\partial x} + (-\sin(\theta)) \frac{\partial ϕ}{\partial r} \cdot \frac{\partial \theta}{\partial x} \\ & + \frac{-\sin(\theta)}{r} \frac{\partial^2 \phi}{\partial \theta^2} \cdot \frac{\partial \theta}{\partial x} + \frac{\partial ϕ}{\partial \theta} \left( \frac{1}{r^2} \sin(\theta) \frac{\partial r}{\partial x} + \frac{1}{r} \left( - \cos(\theta) \right) \frac{\partial \theta}{\partial x} \right)\\ & = \cos^2(\theta) \frac{\partial^2 ϕ}{\partial r} + \frac{\sin^2(\theta)}{r} \frac{\partial ϕ}{\partial r} + \frac{\sin^2 (\theta)}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} + \frac{\partial ϕ}{\partial \theta} \left( \frac{2 \sin(\theta) \cos(\theta)}{r^2} \right) \qquad \square \\ \frac{\partial^2 ϕ}{\partial y^2} & = \sin(\theta) \frac{\partial^2 ϕ}{\partial r^2} \cdot \frac{\partial r}{\partial y} + \cos(\theta) \frac{\partial ϕ}{\partial r} \cdot \frac{\partial \theta}{\partial y} \\ & + \frac{\cos(\theta)}{r} \frac{\partial^2 \phi}{\partial \theta^2} \cdot \frac{\partial \theta}{\partial y} + \frac{\partial ϕ}{\partial \theta} \left( \frac{-1}{r^2} \cos(\theta) \frac{\partial r}{\partial y} + \frac{1}{r} \left( - \sin(\theta) \right) \frac{\partial \theta}{\partial y} \right)\\ & = \sin^2(\theta) \frac{\partial^2 ϕ}{\partial r} + \frac{\cos^2(\theta)}{r} \frac{\partial ϕ}{\partial r} + \frac{\cos^2 (\theta)}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} + \frac{\partial ϕ}{\partial \theta} \left( \frac{-2 \sin(\theta) \cos(\theta)}{r^2} \right) \qquad \square\\ \end{align*} \]

Plugging in our results into our original equation,

\[ \begin{align*} ∇^2 ϕ &= \frac{\partial^2 ϕ}{\partial x^2} + \frac{\partial^2 ϕ}{\partial y^2} + \frac{\partial^2 ϕ}{\partial z^2} \\ &= \frac{\partial^2 ϕ}{\partial r^2} + \frac{1}{r} \frac{\partial ϕ}{\partial r} + \frac{1}{r^2} \frac{\partial^2 ϕ}{\partial \theta^2} + \frac{\partial^2 ϕ}{\partial z^2} \qquad \blacksquare \end{align*} \]