What i want to know in this question is write generalized version of gradient in spherical coordinates for $R^D$.
I know for $3D$ case, \begin{align} \nabla_{R^3} = \partial_r + \frac{1}{r^2} \nabla_{S^2} \end{align} where $S^2$ describes a unit sphere described by two angle variables $(\theta, \varphi)$, thus $\nabla_{S^2} = \sin^2(\theta)d\varphi^2 + d\theta^2$
How about $R^D$? can i generalized this to \begin{align} \nabla_{R^D} = \partial_r + \frac{1}{r^{D-1}} \nabla_{S^{D-1}} \end{align} ?
Take further i want to know the laplacian in terms of $\nabla_{S^2}$ in $3d$ case and its generalization.
Above generalization is wrong!.$ r^2$ is fixed.
Consider 2d case \begin{align} \nabla_{R^2} f = \frac{ \partial f}{\partial r} e_r + \frac{1}{r} \frac{\partial f}{\partial \theta} e_\theta = \frac{ \partial f}{\partial r} \hat{e}_r + \frac{1}{r^2} \frac{\partial f}{\partial \theta} \hat{e}_\theta =\frac{ \partial f}{\partial r} \hat{e}_r + \frac{1}{r^2} \nabla_{S^1}f \end{align} Now for 3d case \begin{align} \nabla_{R^3} f&= \frac{ \partial f}{\partial r} e_r + \frac{1}{r} \frac{\partial f}{\partial \theta} e_\theta + \frac{1}{r\sin(\theta)} \frac{\partial f}{\partial \phi} e_{\phi} \\ &= \frac{ \partial f}{\partial r} \hat{e}_r + \frac{1}{r^2} \frac{\partial f}{\partial \theta} \hat{e}_\theta + \frac{1}{r^2\sin^2(\theta)} \frac{\partial f}{\partial \phi} \hat{e}_{\phi} \\ & = \frac{ \partial f}{\partial r} \hat{e}_r + \frac{1}{r^2} \nabla_{S^2} f \end{align} In this way i can generalized to N dimension case \begin{align} \nabla_{R^N}f=\frac{ \partial f}{\partial r} \hat{e}_r + \frac{1}{r^2} \nabla_{S^{N-1}}f \end{align} And the laplacian can be written as \begin{align} \Delta_{R^N} f = \frac{\partial^2 f}{\partial r^2} + \frac{N-1}{r} \frac{\partial f}{\partial r} + \frac{1}{r^2} \Delta_{S^{N-1}} f \end{align}