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Let $U$ be an open subset of $\mathbb{R}^n$. Let $f\colon U \rightarrow \mathbb{R}$ be a function. Suppose the partial derivative $f_{x_i}$ exists and it is continuous on $U$ for $i = 1, \dots, n$. Is $f$ continuous?

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Yes, this is a standard calculus result. If a function has continuous partial derivatives, then it is actually differentiable, i.e., at every point $x\in U$ you get $$f(x+h) = f(x) + \nabla f(x) h + o(\|h\|).$$ This implies continuity. The way to prove this statement is using the one-dimensional Mean Value Theorem on segments parallel to coordinate axes $n$ times, then continuity of partials shows that all the partial derivatives converge to the partials at $x$ when $h \to 0$. You can find the details in any decent Real Analysis textbook.

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You can go across a bit more.If one has existence of atleast one of partial derivatives $ \ f_x{_i} $ at a point $ c $ in $ U $ and assume that the remaining $ n-1 $ partial derivatives exist in some $ n $-ball $ B(c) $ and are only continuous at $ c $, then you can show that $ f $ is differentiable at $ c $. This holds in general for vector valued functions.The proof is given in Apostol's Mathematical Analysis Pg.357 Ch 12.

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