C^1 implies differentiable

$C^1$ implies differentiable: Let $U\subseteq\mathbb{R}^n$ be open and let $f:U\to\mathbb{R}^m$. Suppose all first-order partial derivatives of $f$ exist on a neighborhood of $a\in U$ and are continuous at $a$ (equivalently, $f\in C^1$ near $a$). Then $f$ is differentiable at $a$.

This theorem provides a practical sufficient condition for differentiability: checking continuity of partial derivatives is often much easier than verifying the definition of differentiability directly.

Proof sketch: For $m=1$, write the increment $f(a+h)-f(a)$ as a telescoping sum along coordinate directions and apply the one-dimensional mean value theorem to each coordinate slice. Continuity of partial derivatives at $a$ shows the error between this increment and the linear map $\nabla f(a)\cdot h$ is $o(\|h\|)$. The vector-valued case follows componentwise.