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En lo que sigue supondremos que $f \in C^2(\mathbb{R}^2 ; \mathbb{R}) .$ Es bien conocido que para $g(r , \theta) = (r \cos{\theta} ,r \sin{\theta})$ la composiciĆ³n $h = f(g)$ esta bien definida y \begin{equation}\begin{pmatrix} \dfrac{\partial h}{\partial r} & \dfrac{\partial h}{\partial \theta} \end{pmatrix} = \begin{pmatrix} \dfrac{\partial f}{\partial x} & \dfrac{\partial f}{\partial y} \end{pmatrix} \begin{pmatrix} \cos{\theta} & - r \sin{\theta} \\ \sin{\theta} & r \cos{\theta} \end{pmatrix},\end{equation} de donde se obtiene \begin{align*}\frac{\partial f}{\partial x} = \cos{\theta} \frac{\partial h}{\partial r} - \frac{\sin{\theta}}{r} \frac{\partial h}{\partial \theta} \\ \\ \frac{\partial f}{\partial y} = \sin{\theta} \frac{\partial h}{\partial r} + \frac{\cos{\theta}}{r} \frac{\partial h}{\partial \theta}\end{align*} De acuerdo con esto $\nabla f$ se puede escribir en coordenadas polares $\nabla_{r, \theta} f$ como $$\nabla_{r, \theta} f = \frac{\partial f}{\partial r} \bm{\hat{r}} + \frac{1}{r} \frac{\partial f}{\partial \theta} \bm{\hat{\theta}}$$ en donde $\bm{\hat{r}} = (\cos{\theta} , \sin{\theta})$ y $\bm{\hat{\theta}} = ( - \sin{\theta} , \cos{\theta})$

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