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I am quite new to total derivatives (up to now I knew the existence of partial derivatives only!). I am dealing some computations that involve second and third-order total derivatives and I have some doubts about what I have done so far (plus I have not found references about high-order total derivatives).

Let $f(x, y(x))$ be a real-valued function, with $x,y\in \mathbb R$. The defintion of total derivative, with respect to $x$, is $$ D_x f(x,y(x))=f_x\{x,y(x)\}+f_y\{x,y(x)\}\frac{\partial y(x)}{\partial x}, $$ where $f_x$ and $f_y$ are the partial derivatives of $f$ with respect to $x$ and $y$, respectively. Now with some kind of "induction", here it is my result for the second-order total derivative (again with respect to $x$)

\begin{eqnarray*} D_x\left[D_x f\{x,y(x)\}\right]&=&f_{xx}\{x,y(x)\}+f_{xy}\{x,y(x)\}\frac{\partial y(x)}{\partial x}+\\ % &+&f_{yx}\{x,y(x)\}\frac{\partial y(x)}{\partial x}+f_{yy}\{x,y(x)\}\left\{\frac{\partial y(x)}{\partial x}\right\}^2+\\ % &+&f_y\{x,y(x)\}\frac{\partial^2 y(x)}{\partial x^2}+f_{yy}\{x,y(x)\}\frac{\partial^2 y(x)}{\partial x\partial y}, \end{eqnarray*} with $f_{xx}$, $f_{yx}$, $f_{xy}$, and $f_{yy}$ the second-order partial derivatives of $f$. Is the above formula correct? If so, I guess that the last term vanishes because $\partial^2 y(x)/\partial x\partial y$ should be $0$.

Thoughts/comments are welcome as well as some references!

  • You shouldn't have the last term. – KittyL Jul 14 '15 at 08:58
  • Thank you! I do not know how to explain, nonetheless I suspected that that term should not be computed at all. – user254434 Jul 14 '15 at 09:09
  • You are right. It should not be computed. The second last term is the second part of the chain rule, which is the derivative of $\frac{\partial y(x)}{\partial x}$. This $\frac{\partial y(x)}{\partial x}$ only involves $x$. So that is it. – KittyL Jul 14 '15 at 09:13
  • Super! Actually, the computations I have to deal with involve $x\in\mathbb R^d,y\in\mathbb R^q,,d,q\geq 1$. I have already developed the second and third-order total derivatives. Now I am more confident about those expressions. Thanks again! – user254434 Jul 14 '15 at 09:33

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