confusion with the meaning of gradients What is the meaning of gradient indeed? Gradient is very common in calculus, analysis but i still not so sure what's gradient after looking at the definition of them. Take the gradient in cartesian and cylindrical coordinate as an example. $\nabla f=f_x i+f_y j+f_z k$ and $\nabla f=\frac{\partial f}{\partial r}\hat r+\frac{\partial f}{r\partial \theta}\hat \theta+\frac{\partial f}{\partial z}k$. I think it is not difficult to see what's going on when the gradient multiply with a infinitesimal vector in cartisean coordinate  which give a linear approximation but for the gradient in cylindrical coordinates, it seems non trivial what's the meaning of it and why the form is not like $\nabla f=\frac{\partial f}{\partial r}\hat r+\frac{\partial f}{\partial \theta}\hat \theta+\frac{\partial f}{\partial z}k$
 A: The gradient of $f$ at $(x,y,z)$  is a linear map from vectors to scalars. It takes a vector $v$ as an input and returns the directional derivative $D_vf(x,y,z)$ as output. So far, this is clean enough. But then we want to write down this linear map using some basis for the space of vectors. As you know from linear algebra, the same linear map has different coefficients when written in different bases. This is all that's going on here.
A: The gradient has the same geometrical meaning regardless of coordinates: it is the vector of the direction of greatest increase of the function.
The reason there's an $\frac{1}{r}$ appearing in cylindrical has to do with what basis of vectors we use to define the gradient.  Indeed, this requires a bit of explanation into how we choose basis vectors in general.
In general, you have your position vector $\mathbf r$ as a function of some coordinates--here, $\mathbf r(r, \theta, z)$.  The partial derivatives of $\mathbf r$ with respect to these coordinates define the basis vectors, also called basis tangent vectors, called $e_r, e_\theta, e_z$.  There's something a little unintuitive when you do this:  as it turns out, $e_\theta = r \hat \theta$.  This is because lines of constant $r,z$ expand as you go out from the origin:  $\theta$ covers more and more distance in the process, so its tangent vector increases in magnitude.  This is also part of the reason we write the volume element in cylindrical as $r \, dr \, d\theta \, dz$.  This is exactly where that $r$ comes from:  the tangent basis vectors determine the volume element.
But there is another basis of vectors, called basis covectors or basis cotangent vectors, denoted $e^r, e^\theta, e^z$, which are defined so that $e^r \cdot e_r = 1$, and similarly for $\theta, z$, while $e^r \cdot e_\theta = 0$, and so on.  There are various ways you can find the cotangent vectors.  If you know how to take gradients in some other coordinate system (like cartesian), then you can find them by considering $\theta = \theta(\mathbf r)$ as a scalar field and computing the gradient $e^\theta = \nabla \theta$.  Again, this does not require you to know how to do this in cylindrical, only in some other coordinate system (Cartesian).
Anyway, it should be clear to you that for $e^\theta \cdot e_\theta = 1$, we must have $e^\theta =\hat \theta/r$ to cancel the $r$ that comes from $e_\theta$.
We use the cotangent basis in the definition of the gradient in all coordinate systems.  So the gradient is really
$$\nabla f = e^r \frac{\partial f}{\partial r} + e^\theta \frac{\partial f}{\partial \theta} + e^z \frac{\partial f}{\partial z}$$
It's only the substitution $e^\theta = \hat \theta/r$ that makes it take the form you usually see.
