Let $E = \mathbb{R}^n$ and $f :E \rightarrow E $ such that $f \in \mathcal{C}^2$. We suppose that $\forall x \in E$, the differential of $f$ is an isometry. Let now $\varphi(x) = \langle\text{d}f(x)h, \text{d}f(x)k\rangle$. Calculate the differential of $\varphi$
I am having a hard time to compute the differential of $\phi$.
My idea would be to express $\varphi$ as the composition of two functions. We have $\varphi(x) = u \circ v(x)$ where $v(x) = (\text{d}f(x)h, \text{d}f(x)k)$ and $u(a,b) = \langle a,b\rangle$. And then simply use the formula of the differential function of composite functions. We have $\text{d}u(a,b)(h_1,h_2) = \langle a,h_2\rangle + \langle h_1,b\rangle$ but I can't understand how to calculate the differential of $v$? Would it be simply:
$$\text{d}v(x)(m) = (\text{d}^2f(x)(h,m),\text{d}^2f(x)(k,m)) $$?
Any help will be appreciated.
Notation: I use the symbol $\partial$ for the Fréchet derivative, the symbol $\partial_k$ for the $k$-th partial derivative and the symbol $d$ for the differential.
Then if $\varphi\in C^1(E,\Bbb R)$ we have that
$$d(\varphi(x))=\sum_{k=1}^n\partial_k \varphi(x)\, dx^k\tag1$$
where $dx^1,\ldots,dx^n$ is the canonical basis of the $1$-forms space over $E$. In this case we have that
$$\begin{align}\partial_k\varphi(x)&=\partial_k\langle \partial f(x)h_1,\partial f(x)h_2\rangle\\&=\langle\partial^2f(x)(h_1,e_k),\partial f(x)h_2\rangle+\langle\partial f(x)h_1,\partial^2 f(x)(h_2,e_k)\rangle\end{align}\tag2$$
where $e_1,\ldots,e_n$ is the canonical basis of $E$. Im not sure if this expression can be simplified.
EDIT:
Oh, I didnt read before... if $\partial f(x)$ is an isometry for all $x$ then $$\varphi(x)=\langle\partial f(x) h_1,\partial f(x) h_2\rangle=\langle h_1,h_2\rangle$$
so $\varphi$ is constant. Consequently $d\varphi=0$.