# How does an absolute value operator interact with partial derivatives?

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.

• 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? Jan 16, 2021 at 13:43
• Yes, we can assume that $f(x,y)$ is continuous and differentiable. Jan 16, 2021 at 14:00
• @J.Galt: what if $f(x,y)=1$ everywhere? Then $g$ is not defined.
– user9464
Jan 16, 2021 at 14:02
• @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. Jan 16, 2021 at 14:06
• @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:-)
– user9464
Jan 16, 2021 at 14:09

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