Use of chain rule in Taylor's Theorem reminder proof

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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?

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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.