I've read that gradient is covariant, which I thought meant that it would use the contravariant basis $e^x$ instead of $e_x$ as below:
$$\nabla f(p) = {df \over dx}e^x + {df \over dx} e^y \phantom{.}|_p$$
$$$$
I tried an example using $f(x)=x^2$ and ${df\over dx} = 2x \phantom{.} e^x$. Suppose we choose to test with position x=1, then the gradient is ${df\over dx}(1) = 2(1) e^x = 2e^x$.
Then letting $u=2x, f(u) = {u^2\over4}, {df\over du} = {u \over 2} e^u$. The same geometric position as x=1 is u=2. So the gradient is ${df\over du}(2) = {2 \over 2}e^u = e^u$.
Having $u=2x$ requires that the $e_u$ basis vector is half of $e_x$, meaning $e^u$ has to double for contravariance.
Below when I plot $2e^x$ and $e^u$ they seem geometrically like the same vector.
$e^x$ and $e" />
Is this reasoning correct?
I guess I am confused as I see on wikipedia it shows the formula using the covariant basis vectors $e_i$ instead, saying that it's dual $df$ uses the contravariant basis vectors ${dx}^i$.
Thanks.