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I wish to take the partial derivative with respect to $y$ of the following expression:

$$g(x,y)=\log{\left|\frac{\partial}{\partial x}f(x,y)\right|}$$

where $f(x,y):\mathbb{R}^2\rightarrow\mathbb{R}$ is some function which depends on $x$ and $y$, and $\frac{\partial}{\partial x}$ denotes the partial derivative with respect to $x$. How does the absolute value operator $|\cdot|$ interact with the partial derivative $\frac{\partial}{\partial y}$ if I want to evaluate $\frac{\partial}{\partial y}g(x,y)$?

I would greatly appreciate any suggestions or references on how to address such situations, also in more general cases.

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  • $\begingroup$ How much do you know about $f$? Is it continuous? Differentiable? It may be that $\frac{\partial}{\partial x}f(x,y)$ exists everywhere but $f$ is not continuous. If so, is it always non-zero? $\endgroup$
    – TonyK
    Jan 16, 2021 at 13:43
  • $\begingroup$ Yes, we can assume that $f(x,y)$ is continuous and differentiable. $\endgroup$
    – J.Galt
    Jan 16, 2021 at 14:00
  • $\begingroup$ @J.Galt: what if $f(x,y)=1$ everywhere? Then $g$ is not defined. $\endgroup$
    – user9464
    Jan 16, 2021 at 14:02
  • $\begingroup$ @mrsamy: For the purpose of this question, you can assume that $f$ is also continuous, differentiable, and not constant. =) I'm not looking for special cases, but for the general rule behind it. $\endgroup$
    – J.Galt
    Jan 16, 2021 at 14:06
  • $\begingroup$ @J.Galt: My point is that that assumption is not sufficient either. Consider the case when $f(x,y)=y^2+4y+e^y$. the function $g$ is still not defined:-) $\endgroup$
    – user9464
    Jan 16, 2021 at 14:09

2 Answers 2

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Assuming that $\frac{\partial }{\partial x}f(x,y)\ne 0$ everywhere, by the chain rule, you need the existence of partial derivative with respect to $y$ for the map $$ y\mapsto |\frac{\partial }{\partial x}f(x,y)| $$

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Write $$g(x,y)=\log|f_x|.$$ Then you have that $$g_y=\frac{1}{|f_x|}\frac{f_x}{|f_x|}f_{xy}=\frac{f_{xy}}{f_x}.$$

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