I know the differential displacement in spherical coordinate as $$dr \cdot \widehat{r}+ r d\theta\cdot\widehat{\theta} + r\sin\theta d\phi\cdot \widehat{\phi}$$. But I can't figure out how the gradient is $\dfrac{\partial}{\partial r}\cdot\widehat{r} + \dfrac{1}{r}\dfrac{\partial}{\partial\theta}\cdot\widehat{\theta} + \dfrac{1}{r\sin\theta}\dfrac{\partial}{\partial \phi}\cdot\widehat{\phi}$. Can anyone show me the deduction please? I am new to this & came across it when was studying Schroedinger's equation in spherical coordinate form.
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1This book is a very good source for your studies. – Giuseppe Negro Jun 16 '15 at 09:17
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This book seems really nice. You can look at this question and my answer to this follow up, too. – Ivo Terek Jun 16 '15 at 09:51
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Following this answer, https://math.stackexchange.com/a/587298/169296, you can use : $$ df = \frac{\partial f}{ \partial r} dr + \frac{\partial f}{ \partial \theta} d \theta + \frac{\partial f}{ \partial \phi} d\phi = \vec{\nabla f }\cdot \vec{dr}$$ Since $\vec{e_r} , \vec{e_\theta} $ and $ \vec{e_\phi}$ is a set of basis vectors you can suppose that $\vec{\nabla f } = \alpha \vec{e_r} + \beta \vec{e_\theta} + \gamma \vec{e_\phi}$.
Then you will find your answer by identifying the coefficients using your expression of $\vec{dr}$.
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Very nice. I would like to point out the similitude between this method and the one that one can find here to compute $\partial_x, \partial_y, \partial_z$ in terms of $\partial_r, \partial_\theta, \partial_\phi$. (Read from "To express partial derivatives with respect to Cartesian axes in terms of partial derivatives of the spherical coordinates ..."). – Giuseppe Negro Jun 16 '15 at 10:04