Differential of gradient on submanifold of Euclidean space I'm currently selfstudying (and new to) differential geometry and I came across something, that I can't make sense of at the moment. Let $M$ be a submanifold of $\mathbb{R}^{n}$ and let $u,f:M\longrightarrow\mathbb{R}$ be a function. The gradient $\nabla^{M}$of $f$ is defined by the property, that $g_{p}(\nabla^{M}f,X)=d_{p}f(X)$, where $p\in M$, $X\in TM$ satisfying $X(p)\in T_{p}M$ and $d_{p}f:T_{p}M\longrightarrow T_{f(p)}\mathbb{R}=\mathbb{R}$. Since the defining property holds for all vector fields $X$, we get that $\nabla^{M}f(p)=g^{ij}d_{p}f$, where $g^{ij}$ is inverse matrix of the matrix representing the first fundamental form.
Now, let $u(x)=\nabla^{M}f(x)$. My questions are:
What is the differential $d_{p}u$ of $u$? Is there, consequently, a meaningful object for something like the "second derivative" aka. the Hessian of $f$?
Thanks in advance.
 A: We can identify $T_x\mathbb{R}$ with $\mathbb{R}$, so we may view the differential of a function $f:M\to\mathbb{R}$ as a covector field $df:M\to T^*M$ defined by
$$
df(X)=X(f)
$$
where $X$ is a vector field.
When working in local coordinates $x^1,\dots,x^d$ (with Einstein summation convention) We can have a frame of partial derivatives $\partial_1\dots,\partial_d$ for $TM$ and a corresponding frame of differentials of the coordinate functions $dx^1,\dots,dx^d$ for $T^*M$, which obey $dx^i(\partial_j)=\delta^i_j$. In coordinates, $df$ is given by the partial derivatives of $f$:
$$
df=\partial_i(f)dx^i
$$
The gradient is then given by
$$
\text{grad} f=g^{ij}\partial_j(f)\partial_i
$$
You can verify that this satisfies the defining property. As for a second derivative, there is such a thing, but it requires a way to differentiate covector fields. This can be done with the covariant derivative. The second covariant derivative (i.e. coveariant Hessian) of a function $f:M\to\mathbb{R}$ is given in coordinates by
$$
\nabla\nabla f=(\partial_i\partial_j f-\Gamma^k_{ij}\partial_k f)dx^idx^j
$$
