The Problem:
Consider the function:
$$ f : \mathbb{R^2} \to \mathbb{R} : (x,y) \mapsto \begin{cases} 2 \sin x + x^2 y \sin\Big(\frac{1}{x^2 + y^4}\Big) \quad &\text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0) \end{cases}$$
In which points of $\mathbb{R^2}$ is $f$ totally differentiable? What is the total derivative in those points?
Attempt:
We can use elementary methods of calculus to find the partial derivatives of $f$ for all $(x,y) \in \mathbb{R^2} \setminus \{(0,0)\}$. In $(0,0)$ we can use the limit definition of partial derivatives and find $(\partial_x f)(0,0) = 2$ and $(\partial_y f)(0,0) = 0$. We can now construct a candidate total derivative for all $(x,y) \in \mathbb{R}^2$, which we name $(df)(x,y)$. It has the $(1 \times 2)$-matrix representation:
$$ (df)(x,y) = \begin{bmatrix} (\partial_x f)(x,y) & (\partial_y f)(x,y) \end{bmatrix} $$
After making the tedious (yet simple) calculations, it is easy to see that both partials are surely continuous everywhere except maybe in $(0,0)$. Because of this, we know that $f$ is differentiable everywhere except maybe in $(0,0)$ and the total derivative is given by $(df)(x,y)$.
It is the point $(0,0)$ that is giving me trouble. After creating some plots with computer software, I have made an educated guess that $f$ is indeed differentiable in $(0,0)$. I have thought of two ways in which I could show this. One way involves showing that $\lim_{(x,y) \to (0,0)} (\partial_x f)(x,y) = 2$ and $\lim_{(x,y)\to(0,0)} (\partial_y f)(x,y) = 0$, which would imply that both partial derivatives are also continuous in $(0,0)$ and hence $f$ is totally differentiable everywhere. The other way involves substituting $(df)(0,0)$ into the definition of the total derivative:
$$ f(x,y) = f(0,0) + (df)(0,0)(x-0,y-0) + o\big(||(x-0,y-0)||\big) \quad \text{as $(x,y) \to (0,0)$}.$$
Simplifying and applying the definition of Landau's little-$o$ notation would then imply that:
\begin{align*} \lim_{(x,y) \to (0,0)} &\frac{|f(x,y) – f(0,0) – (df)(0,0)(x,y)|}{||(x,y)||} \\
= \lim_{(x,y) \to (0,0)} &\frac{\big|2 \sin x + x^2 y \sin \Big(\frac{1}{x^2 + y^4}\Big) – 2x\big|}{\sqrt{x^2 + y^2}} = 0
\end{align*}
Verifying the differentiability of $f$ then comes down to verifying that this limit expression indeed equals $0$. But calculating a limit in two variables is anything but obvious. My attempts to show this limit is not $0$ or does not exist (which would imply $f$ is not differentiable in $(0,0)$) have so far been futile.
Am I on the right path? Have I made any mistakes in the first part of my attempt? Is there any way I could (relatively) easily see that a function is differentiable or not, so that I don't need to resort to plotting the graph? Any tips or extra information would be much appreciated.
Best Answer
Going with the hints in comments... $$\lim_{(x,y) \to (0,0)} \frac{\big|2 \sin x - 2x + x^2 y \sin \big(\frac{1}{x^2 + y^4}\big)\big|}{\sqrt{x^2 + y^4}} \leq \lim_{(x,y) \to (0,0)} \Big|\frac{2 \sin x - 2x}{x}\Big| + \lim_{(x,y) \to (0,0)} \big|xy \sin \big(\frac{1}{x^2 + y^4}\big)\big|$$. By noting that $\frac{\sin x - x}{x} < 0$ for any $x \neq 0$, we can get rid of the absolute value in the first limit. It is then easily seen to be zero. We use the fact that $0 \leq |\sin( \cdot )| \leq 1$ to obtain the following upper bound to the second limit: $\lim_{(x,y) \to (0,0)} |xy|$. While it may be obvious that this bound is $0$, it is in my opinion more rigourous to use the $\varepsilon,\delta$-definition for a quick proof.
Let $\varepsilon > 0$ and take $\delta = \sqrt{\varepsilon} > 0$. Choose $(x,y) \in \mathbb{R}^2$ such that $||(x,y)|| = \sqrt{x^2 + y^2} < \delta$. Then we surely have $|x| < \delta$ and $|y| < \delta$ and, therefore, $|xy| < \delta^2 = \varepsilon$. This proves that $\lim_{(x,y) \to (0,0)} |xy| = 0$.
We have now bounded the desired limit above by 0. But since the expression inside the limit itself is always greater than or equal to 0, the limit must be 0. This proves that $f$ is totally differentiable in $(0,0)$.