If $f : \mathbb{R}^{n} \to \mathbb{R}$ has continuous second partial derivatives, then $\forall i, j $

$$\frac{\partial^2 f(x_1, \dots, x_n)}{\partial x_i\, \partial x_j} = \frac{\partial^2 f(x_1, \dots, x_n)}{\partial x_j\, \partial x_i}.\,\! $$

Assume the function $g(x_1,\dots,x_n)$ has continuous second partial derivatives, can we conclude the following ?

$$\frac{\partial}{\partial g} \left( \frac{\partial g}{\partial x_i} \right) = \frac{\partial}{\partial x_i} \left( \frac{\partial g}{\partial g} \right) = \frac{\partial}{\partial x_i} (1) = 0$$


There is no meaning in $\frac{\partial}{\partial g}$. We always take the partial derivatives w.r.t. some of the variables $x_i$.

  • $\begingroup$ The differential here is defined in the meaning of the Fréchet derivative. $\endgroup$ – Christine Darcoux Jun 3 '14 at 23:16

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