Directional derivative, gradient and metric Considering the general expression of gradient with the directional derivative operator on $f$ function along $\vec{v}$ vector :
$$df(v)=\langle\text{grad}(f),v\rangle = g^{ij}\partial_{i} f v_{j}=\partial_{i} f v^{i}$$
taking $\partial_{i} = \dfrac{\partial}{\partial x^{i}}$ (with $x^{i}$ contravariant coordinates).
Can I write also :
$$df(v)=\langle\text{grad}(f),v\rangle = \partial^{i} f v_{i}$$
with $\partial^{i} = g^{ij} \partial_{j} =\dfrac{\partial}{\partial x_{i}}$ where $x_{i}$ are covariant coordinates ??
i.e, I don't know if I can raise up the index of $\partial_{j}$ multiplying it by $g^{ij}$ while defining $\partial^{i} = \dfrac{\partial}{\partial x_{i}}$ ?
Regards
EDIT 1 : it may be that I do confusions between covariant/contravariant coordinates of a vector and curvilinear coordinates (curviliear coordinates are always contravariant, aren't they ?)
EDIT 2 : question transferred on https://math.stackexchange.com/questions/2823029/directional-derivative-gradient-and-metric

If you are in a flat Euclidean space $\Bbb R^n$ and you are using the euclidean metric then the metric tensor is $g_{ij}=\delta_{ij}$ as well
$g^{ij}=\delta^{ij}$. So the law of raising indexes gives that $\partial^i=\partial_i$.
