# Converting potential gradient into polar coordinate

In polar coordinate,

$$\nabla U = \frac{\partial U}{\partial r}\hat{\mathbf{r}} + \frac{1}{r}\frac{\partial U}{\partial \theta}\hat{\mathbf{\theta}} .$$

I tried to use chain rule to get this result, but what I got from $\nabla U = \frac{\partial U}{\partial x}\hat{\mathbf{x}}+\frac{\partial U}{\partial y}\hat{\mathbf{y}}$ is some gibberish. Can anyone explain in detail how to get this result?

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Given that $$r = \sqrt{x^2 + y^2},\qquad \theta = \arctan\left(\frac{y}{x}\right)$$ and $U = U(x,y)$, using the chain rule in two variables, we have $$\frac{\partial U}{\partial x} = \frac{\partial U}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial U}{\partial \theta}\frac{\partial \theta}{\partial x}$$ $$\frac{\partial U}{\partial y} = \frac{\partial U}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial U}{\partial \theta}\frac{\partial \theta}{\partial y}$$ Now, $$\frac{\partial r}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^2+y^2} = \frac{x}{\sqrt{x^2 + y^2}} = \cos \theta$$ $$\frac{\partial r}{\partial y} = \frac{\partial}{\partial y} \sqrt{x^2+y^2} = \frac{y}{\sqrt{x^2 + y^2}} = \sin \theta$$ and $$\frac{\partial \theta}{\partial x} = \frac{\partial}{\partial x} \arctan \left(\frac{y}{x}\right) = -\frac{y}{x^2 + y^2} = -\frac{\sin \theta}{r}$$ $$\frac{\partial \theta}{\partial y} = \frac{\partial}{\partial y} \arctan \left(\frac{y}{x}\right) = \frac{y}{x^2 + y^2} = \frac{\cos \theta}{r}$$ hence $$\nabla U = U_r \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix} + \frac{1}{r} U_\theta \begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix}$$ Lastly, the unitary vector in the radial direction $\hat{r}$ is $$\hat{\textbf{r}} = \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}$$ and due orthogonality $$\hat{\theta} = \begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix}$$ which leads to the desired form $$\nabla U = U_r \hat{\textbf{r}} + \frac{1}{r} U_\theta \hat{\theta}$$