# Does chain rule require continuously differentiability?

Recently I read the book Advanced Calculus written by Fitzpatrick. The Theorem 15.34 tells that If $F:\mathbb R^n\to\mathbb R^m$ is CONTINUOUSLY differentiable (all partial derivatives exist and continuous) and $g:\mathbb R^m\to\mathbb R$ is also CONTINUOUSLY differentiable, then $\frac{\partial}{\partial x_i}(g\circ F)(x)=\sum^m_{j=1}D_jg(F(x))\frac{\partial F_j}{\partial x_i}(x)$ and $g\circ F$ is also continuously differentiable.

I know that continuously differentiable (all partial derivatives exist and continuous) implies differentiable. But a function is differentiable may not indicate that its partial derivatives are continuous. I wonder that whether the chain rule can be applied if it is differentiable but do not have continuous derivatives.

p.s. The proof in the book contains $\lim_{h\to0}\nabla g(x+h)=\nabla g(x)$ which needs continuity of $\nabla g$

• The chain rule only needs differentiability at the relevant points. Continuity of the derivative is not required. Also, I prefer the notation $D (g \circ F)(x) = D g(F(x))(D F(x))$. – copper.hat Jul 25 '14 at 4:39
• But differentiability here means total differentiability. At least, partial differentiability alone is not sufficient. – PhoemueX Jul 25 '14 at 4:43
• @PhoemueX: Of course, the function defined on the plane to be one everywhere except zero on the axes has partials at $(0,0)$ but is not even continuous. – copper.hat Jul 25 '14 at 6:35