Derivering ger \[ \begin{array}{cc} \dfrac{\partial r}{\partial x}= & \dfrac{\partial}{\partial x}\sqrt{x^2+y^2}= \dfrac{x}{\sqrt{x^2+y^2}}=\cos\theta, \\ \dfrac{\partial r}{\partial y}= & \dfrac{\partial}{\partial y}\sqrt{x^2+y^2}= \dfrac{y}{\sqrt{x^2+y^2}}=\sin\theta, \\ \dfrac{\partial \theta}{\partial x}= & \dfrac{\partial}{\partial x}\arctan\dfrac{y}{x}= \dfrac{-y}{x^2+y^2}=-\dfrac{\sin\theta}{r}, \\ \dfrac{\partial \theta}{\partial y}= & \dfrac{\partial}{\partial y}\arctan\dfrac{y}{x}= \dfrac{x}{x^2+y^2} =\dfrac{\cos\theta}{r}, \end{array} \]vilket ger matrisen\[ \left( \begin{array}{cc} \cos\theta & -\dfrac{\sin\theta}{r} \\ \sin\theta & \dfrac{\cos\theta}{r} \end{array} \right). \]Enkel matrismultiplikation ger att\[ \left( \begin{array}{cc} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{array} \right) \left( \begin{array}{cc} \cos\theta & -\dfrac{\sin\theta}{r} \\ \sin\theta & \dfrac{\cos\theta}{r} \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)\]och\[ \left( \begin{array}{cc} \cos\theta & -\dfrac{\sin\theta}{r} \\ \sin\theta & \dfrac{\cos\theta}{r} \end{array} \right) \left( \begin{array}{cc} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right).\]