# Differentiability and continuity while partials have different conditions

The relevant things I read and will discuss are in this snapshot (from Folland Advanced calculus, and Wolfram Alpha, and this answer by zhw for an old question).

Also, let me add two other links which I read that might be helpful: this question Continuity of one partial derivative implies differentiability, in particular this answer.

Now First question I have is without knowing $$f$$ is differentiable or continuous how do we use mean value theorem. Do We think as $$f_1$$ and $$f_2$$ are the functions of only $$x$$ and $$y$$ respectively?? Also proving this Folland uses mean value theorem Does it really necessary to use the Mean Value Theorem? Here's a picture of my proof.

1. Second question is how the functions in the picture are bounded(wolfram alpha results)? It says in question 8 that example 7 provides examples for function that satisfies this (makes $$f$$ continuous but not differentiable). The functions $$f_1$$ and $$f_2$$ are not continuous (everywhere) but they are if we take $$x$$ and $$y$$ constant respectively then they are bounded because everywhere cont functions takes bounded to bounded

2. question is In proof of zhw and mine: what happens if we remove the fact all partials are continuous? What causes it to fail in my proof? I think removing continuity causes proof to fail at where I write "by continuity". However, I don't quite see where it fails at zhw's answer. If we don't use continuity we can say "it exists" for $$f_x$$ (derivative of $$f$$ with respect to $$x$$) as well.

3. question is I solved it without using Mean Value Theorem. Again, I just can't seem to understand why do we need it because if we can prove without using it then there is no need why everyone uses it then :D

I know it is a long question but they all connected Sorry for bad format I don't know how to use fancy symbols etc. Thank you for your answers already.

• Welcome to MSE. Please learn to use MathJax to properly format math expressions. The URL link to any answer post can be obtained by clicking the tiny "share" at the bottom left of that post. – Lee David Chung Lin Apr 23 at 21:57