C^1 functions are strictly differentiable

Statement

Let $U \subseteq \mathbb{R}^n$ be open, and let $f \colon U \to \mathbb{R}^m$ be of class C^1 on $U$. Then $f$ is strictly differentiable at each point $a \in U$, with linear map $Df(a)$.

Remarks

Proof: let $\varepsilon > 0$ and choose $r > 0$ with $\|Df(z) - Df(a)\| \le \varepsilon$ for $\|z - a\| < r$. Apply the mean value inequality to $g(z) = f(z) - Df(a) z$ on the ball of radius $r$. This gives $\|f(x) - f(y) - Df(a)(x - y)\| \le \varepsilon \|x - y\|$ for $x, y$ in that ball. For a function of one variable on an open interval, strict differentiability at each point is equivalent to being of class $C^1$.

Explanations

L1 human

If the derivative changes continuously, then near $a$ it stays close to $Df(a)$. So every chord of the graph near $a$ has a slope close to $Df(a)$, which is what strict differentiability asks.