Integral of directional derivative.

I have just learned about a proof for why the gradient "shows the direction of steepest ascent" involving:

$$\partial_{v_0} f(x_0) = (f \circ \gamma)^{\prime}(t_0) = \langle\nabla f(x_0), v_0\rangle$$

with: $$\gamma (t_0) = x_o, \gamma^{\prime}(t_0)=v_0$$

which I now understand. A little later in the script, it is presented that:

$$f(x_1) - f(x_0) = \int^b_a \langle \nabla f (\gamma(t)), \gamma^{\prime}(t)\rangle dt$$.

for every $$C^1$$-curve $$\gamma:[a,b] \rightarrow U \subset \mathbb{R}^n$$ with $$\gamma(a) = x_0, \gamma(b) = x_1$$, which is without proof or definition. Does anybody have an intuition or explanation as to why this holds true or maybe just a link to where I can read more?

• The first equation tells you what you want to know, if you recall that the directional derivative in direction $v_0$ will therefore be greatest when $v_0$ is in the direction of $\nabla f(x_0)$. (Recall that $u\cdot v\le |u||v|$, with equality holding when $u$ and $v$ point in the same direction.) Sep 22 '19 at 1:13
• Oh, I remember, thanks. Sep 22 '19 at 2:59
• @TedShifrin But whats not so obvious to me, is why $f(x_1)-f(x_0) = \int^1_0 df(x_0 + t(x_1-x_0)dt \cdot (x_1-x_0)$ when $[x_0,x_1] = \{ x_0 + t(x_1-x_0): 0 \leq t \leq 1\} \in U \subset \mathbb{R}^n$ would hold true. Specifically, I don't understand why we multiply with $(x_1-x_0)$ in the end. Could you help me with that? Sep 22 '19 at 3:04
• Well, as the answer below says, you're just integrating $(f\circ\gamma)'(t)$ from $0$ to $1$ and using the chain rule to break that derivative down (and then the FTC, as I said below). You can also think of this as a line integral of $\nabla f$ along the path from $x_0$ to $x_1$ (which might be a line segment or might be a more general path). Then it's the FTC for line integrals. (By the way, if you're interested in proofs, you might find my YouTube videos for multivariable calculus/analysis and linear algebra interesting or helpful. They're linked in my profile.) Sep 22 '19 at 3:41

By the chain rule, the derivative of $$t \mapsto f(\gamma(t))$$ is $$f^\prime(\gamma(t)) \circ \gamma^\prime(t)=\langle \nabla f (\gamma(t)), \gamma^{\prime}(t)\rangle$$