TITLE: fundamental theorem of calculus relation
QUESTION [5 upvotes]: I sent this question last week but it turns out that I had an error in the formula that I was trying to understand. So, here goes a second try. I am reading a proof in a convex optimization book by Nesterov and at the beginning of the proof it says the following:
for any $x,y \in \mathbb{R}^n$, we have
$f'(y) = f'(x) + \int_0^1 f''\left(x + \tau(y-x)\right)(y-x)~d\tau$
The earlier answer referred to the fundamental theorem of calculus but I read about the fundamental theorem of calculus and I'm still not seeing it. Thanks a lot.

REPLY [6 votes]: Consider the function
$$
g(\tau):=f'(x+\tau (y-x)).
$$
Then by chain rule
$$
g'(\tau)=f''(x+\tau (y-x))(y-x).
$$
Now the fundamental theorem of calculus yields
$$
g(1)-g(0)=\int_0^1 g'(\tau)d\tau
$$
which can be rewritten
$$
f'(y)-f'(x)=\int_0^1 f''(x+\tau (y-x))(y-x)d\tau.
$$