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Let's consider a domain $E \in \mathbb{R}^d$ and a function $f(x,y): E \times E \to \mathbb{R}$. Suppose $f \in C^2(E \times E)$. If we define a function $g$ on $E$ by $g(x):=f(x,x)$, is it true that $g \in C^2(E)$?

Thank you very much!

user7762
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Yes, it's true. The derivatives of $g$ can be computed by the chain rule.

John Hughes
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  • Yes I agree that the partial derivatives of g can be computed by the chain rule. But the existence of partial derivatives does not imply differentiability. – user7762 Jan 21 '14 at 03:39
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    It's true that the chain rule lets you compute partials...but it also lets you compute the whole derivative. If $g = f \circ h$ (and $f$ and $h$ are differentiable), then $g$ is too, and $g'(x) = f'(h(x)) \cdot h'(x)$. (Milnor's Topology from the Differentiable Viewpoint has a wonderfully concise description of this in the first chapter). In your case, $g = f \circ h$ where $h:E \to E \times E : x \mapsto (x, x)$ is the diagonal map, which is certainly differentiable. So you're done. – John Hughes Jan 21 '14 at 04:28