Use of chain rule in Taylor's Theorem reminder proof In the book "An Introduction to Manifolds" by Loring Tu, in the proof of the Taylor's Theorem with remainder the chain rule id used in the following way:
Let $f$ be a smooth function on an open subset $U$ of ${R}^n$ star-shaped with respwect to apoint $p=(p^1,...,p^n)$ in $U$. Using the chain rule 
$$ \frac{d}{dt}f(p+t(x-p))=\sum_{i=1}^n(x^i-p^i)\frac{\partial}{\partial x^i} f(p+t(x-p)) .$$
However, it seems to me that if we define $z^i=p^i+t(x^i-p^i)$, what we should actually have 
$$ \frac{d}{dt}f(p+t(x-p))=\sum_{i=1}^n(x^i-p^i)\frac{\partial}{\partial z^i} f(p+t(x-p)) .$$
And since $\frac{\partial}{\partial x^i} z^i=t $, I don't understand how these two formulas can be the same. Once in a while I encounter things like this and it always baffles me. 
I have seem a similar question here but it is claimed that they are just using the variable $x$ for two different things. But why would you do this? 
It is so ambiguous,especially in this situation where it would mean using $x^i$ for two different things in the $same$ expression! And furthermore they were careful enough to write the argument of the function explicitly. 
So is it really the case that they are just using $x$ for different things? If so, when I encounter things like this can I safely assume they are just being ambiguous, i.e. is this common practice?

You are right. It is the same notation for two different things. I guess they first define $f(x_1, x_2,\ldots, x_n)$, so $\frac{\partial f}{\partial x_i}$ denotes the $i$-th partial derivative of $f$, which is then evaluates at $p+t(x-p)$ where $x$ is the name for a variable. Quite common I would say.
