$f: M \to \mathbb R$, smooth, $\varphi_t$ 1 parameter group of diffeomorphisms, then $df(\varphi_t(x))/dt = 1$ I will first introduce some preliminaries and notation. The question is highlighted below.
$M$ smooth manifold of dimension $N$, $f:M \to \mathbb R$ smooth.
Fact: a given vector field $\underline v$ on $M$ which vanishes outside a compact subset of $M$ generates a $1$-parameter group of diffeomorphisms $\varphi_t(x)$ which is the solution of
$$
\frac{d \eta}{dt} = \underline v(\eta(t)), \quad \eta(0) = x.
$$
We can construct a Riemannian metric on $M$, which gives a notion of inner product in the tangent spaces and allows for the definition of the gradient vector:
$$
\langle v, \text{grad} f \rangle = df(v).
$$
Consider $f^{-1}([a, b])$ a compact set without critical points of $f$. We can define a map $\rho: M \to \mathbb R$ given by
$$
\rho = \frac{1}{\langle \text{grad} f, \text{grad} f \rangle}
$$
in $f^{-1}([a, b])$ and $0$ outside a compact neighborhood of $f^{-1}([a, b])$. Then $\underline v = \rho \text{grad}f$ generates a $1$-parameter group of diffeomorphisms.

The book presents the following computation:
$$
\frac{df(\varphi_t(x))}{dt} = \left\langle \frac{d\varphi_t(x)}{dt}, \text{grad}f \right \rangle = \langle x, \text{grad}f \rangle = 1.
$$
Clearly there is a typo in the third member of the chain of inequalities. In order for us to get $1$ in the right-hand side, the third member should be $\rho\langle \text{grad}f, \text{grad}f \rangle$. However, as I understand, we only have
$$
\frac{d\varphi_t(x)}{dt} = \rho \text{grad}f(\varphi_t(x)),
$$
and we would obtain
$$
\rho \langle \text{grad}f(\varphi_t(x)), \text{grad}f \rangle.
$$
Please, what am I missing?

This is in pages 64/65 of Damascelli and Pacella's Morse Index of Solutions of Nonlinear Elliptic Equations, and appears in the proof that we can deform a sublevel into another if there is not critical level between them.
Thanks in advance.
 A: The problem is that the author has chosen not to write the variables in the equality. This is more pleasant to read but we can be confused while theses equalities occure. Conversely, writting everything is sometimes harsh.
I will denote by $g$ the Riemannian metric of $M$: if $u$ and $v$ are tangent to $M$ and $p$, then their scalar product is denoted by $g_p(u,v)$ instead of $\langle u, v\rangle$, so we keep the information on where do these vectors live. The gradient of a function $h$ is then defined thanks to the relation
$$
\forall v \in T_pM, \mathrm{d}h(p)\cdot v = g_p\left(\mathrm{grad}h(p),v \right)
$$
If $f : M \to \mathbb{R}$ is smooth, and if $v = \rho \mathrm{grad}f$ is the vector field you defined above, then
\begin{align}
\frac{\mathrm{d} f\left(\varphi_t(x)\right)}{\mathrm{d}t} &= \mathrm{d}f\left(\varphi_t(x)\right)\cdot \left(\frac{\mathrm{d}}{\mathrm{dt}}\varphi_t(x)\right) \\
&=g_{\varphi_t(x)}\left(\mathrm{grad}f(\varphi_t(x)), \frac{\mathrm{d}}{\mathrm{d}t}\varphi_t(x) \right) \\
&= g_{\varphi_t(x)}\left(\mathrm{grad}f(\varphi_t(x),\rho(\varphi_t(x))\mathrm{grad}f(\varphi_t(x)) \right) \\
&= \rho(\varphi_t(x)) g_{\varphi_t(x)}\left(\mathrm{grad}f(\varphi_t(x)),\mathrm{grad}f(\varphi_t(x)) \right) \\
&=1.
\end{align}
As you can see, the notations here are a little bit unpleasant, but we find the right proof.
