Taylor's theorem with remainder In Loring Tu's Introdution to Manifolds pg. 6, it was said that if $f$ is a $C^\infty$ function, then $$g_i(x)=\int^1_0 \frac{\partial f}{\partial x^i}(p+t(x-p))dt$$
is $C^\infty$.  Tu used $x$ in two different ways, both as a variable and as a point.
Why is $g_i(x) $ $C^\infty$?
 A: While I agree that Tu would have been wiser to have called $x$ by a different letter, it is not accurate to say he is using it in two different ways.
$x^i$ is not some reference to $x$. It is an entirely separate variable. In particular, it refers to the $i$-th coordinate on $\Bbb R^n$ (where $n$ is the dimension of the domain of $f$, which you failed to include (and you should have known better, the domain of $f$ is highly relevant information and it is impolite to make people guess what you are talking about).
So what it is saying is, first differentiate $f$ with respect to its $i$-th coordinate, then evaluate the partial derivative at $p + t(x - p)$, and integrate that function with respect to $t$.
Because $f$ is $C^\infty, \frac{\partial f}{\partial x^i}$ is $C^\infty$. That is what $C^\infty$ means: you can differentiate infinitely many times. And integration is a very friendly operation. It makes functions more regular, not less, so it doesn't break that smoothness. Check the theorems  on the interchange of integrals and derivatives in your favorite Real Analysis text.
