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Let $g(t)=(x(t),y(t))$ and suppose that $g'(t)$ exist in $t=t_0$. By the chain rule, if the partial derivatives of a function $f(x,y)$ are continuous in an open neighborhood of $g(t_0)=P$, then $$ (f\circ g)'(t_0)=f_x(P)x'(t_0)+f_y(P)y'(t_0). $$ Are the partial derivatives $f_x$ and $f_y$ must be continuous in $P$ or there exist a version of the chain rule with weaker conditions?

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Let $A\subset\mathbb{R}^{n},\ B\subset\mathbb{R}^{m}$ open sets and let $f:A\to\mathbb{R}^{m},\ g:B\to\mathbb{R^{l}}$ be functions such that:

  1. $f$ is differentiable at $a\in A$
  2. $g$ is differentiable at $f(a)$
  3. $f(A)\subseteq B$

Then, $g_{o}f$ is differentiable at $a$ and $(g_{o}f)'(a)=g'(f(a))f'(a)$.

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  • $\begingroup$ This is not what the OP was asking for. The OP wanted to know if $f$ differentiable implies that its partials are continuous. $\endgroup$ Commented Mar 21, 2017 at 16:56
  • $\begingroup$ are you sure? my english isnt good but: "Are the partial derivatives $f_{x}$ and $f_{y}$ must be continuous in $P$ or there exist a version of the chain rule with weaker conditions?" Since continuous partial drivatives implies differentiability, the theorem i wrote is stronger $\endgroup$ Commented Mar 21, 2017 at 17:19
  • $\begingroup$ But differentiability does not implies that the partial derivatives are continuous... $\endgroup$
    – boaz
    Commented Mar 22, 2017 at 7:45
  • $\begingroup$ read carefully! $\endgroup$ Commented Mar 22, 2017 at 18:42

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