How to expand out this partial derivative (breaking time calculation)? In my professor's notes, he finds $$\frac{\partial }{\partial x_i} \vec{c}(f(\vec{x_0}(t, x_1, \dots, x_d)))t$$
where $t, x_1, x_2, \dots x_d$ are independent variables, $\vec{x_0} = [x_{01}, x_{02}, \dots, x_{0d}]$ is a vector-valued function from $\mathbb{R}^{d+1}$ to $\mathbb{R}^d$, $f$ is a real-valued function on $\mathbb{R}^d$ and $\vec{c}$ is a vector-valued function from $\mathbb{R}$ to $\mathbb{R}^d$.
He evaluates this to be
$$t \times \big(\vec{c}\ '(f(\vec{x_0})) \cdot \nabla_xf(\vec{x_0})\big) \times \frac{\partial}{\partial x_i} \vec{x_0} $$
Where $\cdot$ denotes the dot product and $\times$ denotes multiplication of a scalar and a vector (or normal multiplication, in the case of the first $\times$ symbol). $\frac{\partial}{\partial x_i} \vec{x_0}$ is taken to represent the vector $[\frac{\partial}{\partial x_i} x_{01}, \frac{\partial}{\partial x_i} x_{02}, \dots, \frac{\partial}{\partial x_i} x_{0d}]$.
When I do the working, I find that I get that this derivative works out to be
$$t \times \vec{c}\ '(f(\vec{x_0})) \times \bigg(\nabla_xf(\vec{x_0}) \cdot \frac{\partial}{\partial x_i} \vec{x_0}\bigg) $$
My question is as follows: are these two equivalent? If not, is there a problem with my working? The context is in trying to compute the breaking time of a quasilinear partial differential equation. My working is as follows:
$$\frac{\partial }{\partial x_i} \vec{c}(f(\vec{x_0}(t, x_1, \dots, x_d)))t = t\frac{\partial }{\partial x_i} \vec{c}(f(\vec{x_0}(t, x_1, \dots, x_d))) $$
$$= t \times \bigg[\frac{\partial }{\partial x_i} c_1, \dots, \frac{\partial }{\partial x_i} c_d\bigg]$$
$$= t \times \bigg[c_1' \frac{\partial }{\partial x_i} f(\vec{x_0}), \dots, c_d' \frac{\partial }{\partial x_i} f(\vec{x_0})\bigg]$$
$$= t \times \bigg( \vec{c}\ ' \times \frac{\partial }{\partial x_i} f(\vec{x_0})\bigg)$$
$$= t \times \bigg(\vec{c}\ ' \times \bigg(\frac{\partial f}{\partial x_{01}} \times \frac{\partial x_{01}}{\partial x_i} + \dots + \frac{\partial f}{\partial x_{0d}} \times \frac{\partial x_{0d}}{\partial x_i}\bigg)\bigg)$$
$$= t \times \vec{c}\ ' \times \bigg(\nabla_x f \cdot \frac{\partial \vec{x_0}}{\partial x_{i}}\bigg)$$
$$ $$
If my working is correct, would it be possible to compute and then factorise the derivative $\frac{\partial }{\partial x_i} \big( \vec{c}(f(\vec{x_0}))t + \vec{x_0}\big)$ to get
$\big(t \times \vec{c}(f(\vec{x_0})) \cdot \nabla_x f(\vec{x_0}) + 1\big) \times \frac{\partial \vec{x_0}}{\partial x_{i}}$?
 A: $\newcommand\R{\mathbb R}$
My answer coincides with yours. The notes may have a mistake.
Here I repeat the calculation in another notation, just to show another (although almost the same) way of doing it.
Ok so the situation is
$$\R^d\xleftarrow{c}\R\xleftarrow{f}\R^d\xleftarrow{x_0}\R^{d+1}$$
Fix $t\in\R$ and call $g:\R^d\to\R^d$ the function given by $g(x_1,\dots,x_d)=x_0(t,x_1,\dots,x_n)$.
Then you want to differentiate the composition
$$\R^d\xleftarrow{c}\R\xleftarrow{f}\R^d\xleftarrow{g}\R^d.$$
(I will ignore the multiplication by $t$ since it is held constant all the time).
By the chain rule, you have
$$\partial_i(c\circ f\circ g)
= (c'\circ f\circ g)\partial_i(f\circ g)
= (c'\circ f\circ g)\sum_{j=1}^n(\partial_jf\circ g)\partial_i g_j.$$
When evaluating at $x=(x_1,\dots,x_n)$, you get
$$\left.\frac{\partial (c\circ f\circ g) }{\partial x_i}\right|_x
= c'(f(g(x)))\sum_{j=1}^n\left.\frac{\partial f}{\partial x_j}\right|_{g(x)}
\left.\frac{\partial g_j}{\partial x_i}\right|_x.$$
Of course, $g_j=x_{0j}$ for $j=1,\dots,n$, so that is the same you got.
Edit:
The version of the chain rule I'm using is that, for a composition
$$\R\xleftarrow\psi\R^m\xleftarrow\phi\R^n,$$ you have
$$\partial_i(\psi \circ\phi)
= \sum_{j=1}^m(\partial_j\psi\circ \phi )\,\partial_i\phi_j$$
for every $i=1,\dots,n$. This is usually writtetn in terms of matrices and gradient vectors, but I find this version more appealing in cases like these.
