# Does continuity of all partial derivatives imply continuity of all directional derivatives?

Let $$X \subset \mathbb R^n$$, $$f:X\to\mathbb R^m$$, $$x_0\in X$$

Assumption: All partial derivatives of f at $$x_0$$ exist and are continuous

$$\Rightarrow$$ f is differentiable at $$x_0$$.

$$\Rightarrow D_vf(x_0)=\nabla f(x_0)\cdot v$$ (assuming $$m=1$$ for simplicity)

Which means that all directional derivatives of f at $$x_0$$ can be expressed as linear combination of the (continuous) partial derivatives of f at $$x_0$$

Therefore these directional derivatives also have to be continuous. (*)

Is the conclusion correct? (Or why not?) And if yes, is my proof correct? (Or why not?)

(*) I'm implicitly assuming that differentiability at $$x_0$$ implies differentiability at all points around $$x_0$$ if they are close enough. Only with this assumption I can conclude that the partial derivatives are defined around $$x_0$$ and therefore ask if they are continuous around $$x_0$$ or not.

I hope you can follow my thoughts. Else just ask for clarifications, it's my first question. Thank you for your help :)

• I believe continuity of partial derivatives implies existence of the derivative and hence all directional derivatives. You can see Spivak's 'Calculus on Manifolds' theorem 2.8 – NL1992 Apr 11 at 19:50
• Yes, I understand that: continuous partial derivatives => differentiability <=> differential exists => directional derivatives exist. But are the directional derivatives continuous? – Stefan Apr 11 at 19:52
• If you assume f has continuous partial derivatives at A open, then it has a continuous differential in A, hence the directional derivatives are also continuous. – NL1992 Apr 11 at 20:08
• In your proof, to be safe, I would've used the fact f is continuously differentiable. – NL1992 Apr 11 at 20:10
• You need to tell us more about $X.$ – zhw. Apr 11 at 20:53