Let $f: B \to \mathbb{R}$ be a $C^\infty$ function on an open ball $B := B_r(a) \subseteq \mathbb{R}^n$. I want to show that there exist $C^\infty$ functions $g_1, ..., g_n: B \to \mathbb{R}$ with
(1) $f(x) = f(a) + \sum_{i=1}^n (x_i - a_i)g_i(x)$ for all $x \in B$,
(2) $g_i(a) = \frac{\partial f}{\partial x_i}(a)$.
I must admit that I haven't come too far myself. My idea is to look at $f$ at the path $\{a + t(x - a): t \in [0, 1]\}$, and then somehow decompose it into it's coordinates show the existence (or concretely define) the $g_i's$, and then show that they also satisfy (2)? I had difficulties in concretely doing that, though.
Also, it's relatively abstract to imagine what I'm working with here, as I know nothing about how $f$ looks like, other than that it's smooth. Is there maybe any intuition to these $g_i's$ or to this exercise in general that makes it somewhat easier?
Define $g(t) = f(a+ t(x-a)), t \in [0,1].$ Then
$$g(1) - g(0) = \int_0^1 g'(t)\, dt = \int_0^1 \nabla f (a+t(x-a))\cdot (x-a)\, dt $$ $$= \int_0^1 \sum_{k=1}^{n} D_kf(a+t(x-a))(x_k-a_k)\, dt = \sum_{k=1}^{n}(x_k-a_k)\left(\int_0^1 D_k f(a+t(x-a)\, dt\right ).$$
Setting $g_k(x) = \int_0^1 D_k f(a+t(x-a))\, dt,$ we have the result. Just check that each $g_k \in C^\infty$ by differentiation through the integral sign.