Differentiability of 2 variable functions

Question

In class, we were taught that if a function $z=f(x,y)$ is differentiable at $(x_0,y_0)$, then first order partial derivatives $f_x(x_0,y_0)$ and $f_y(x_0,y_0)$ exist at that point, and it satisfies the "increment theorem" i.e. $\Delta z=f_x(x_0,y_0)\Delta x+f_y(x_0,y_0)\Delta y + \epsilon \Delta x+\eta\Delta y$ where $\epsilon \to 0, \, \eta\to0 $ as $\Delta x \to 0, \, \Delta y \to 0 $ which, as I understand, intuitively expresses that the function can be approximated locally by a linear function. However, as a computational check for differentiability, we were taught this equation:

$$\lim_{(h,k) \to (0,0)} \dfrac{f(x+h,y+k)-f_x(x_0,y_0)-f_y(x_0,y_0)-f(x_0,y_0)}{\sqrt{h^2+k^2}}=0$$

How do you go from the increment theorem to the above limit? I'm not very familiar with typical $\epsilon-\delta$ arguments (I'm an engineering student). Does this rely on them? I thought this might rely on the conversion of some multivariable limit from Cartesian to polar coordinates, but I can't figure out the details if that is the case.

Furthermore, I did some reading and apparently, there are two kinds of differentiability; weak (Gateaux) and strong (Frechet) where the latter implies the former. And, for certain functions where the first-order partial derivatives are discontinuous (like $f(x,y)=\dfrac{x^5+y^3}{x^4+y^2}$ or $f(x,y)=\dfrac{x^2y^3}{x^4+y^4}$), they are said to be "Gateaux differentiable" but not Frechet differentiable.

How does the above definition of differentiability fit into this framework? I think it is the definition of Gateaux differentiability simplified for two variable functions but I'm not sure. Also, the continuity of partial derivatives implies differentiability but not vice versa. Is there any proof of this claim? What are the common counterexamples i.e. differentiable functions with discontinuous first-order partial derivatives?

Lastly, how does differentiability in the Frechet sense, Gateaux sense, and in the sense of the above equation, cement the relationship between the directional derivative and gradient? We learned in class that the fundamental definition of the directional derivative is $$D_{\hat{u}}f(P)=\lim_{h\to0} \dfrac{f(x_0+hu_x,y_0+hu_y)-f(x_0,y_0)}{h}$$ For differentiable functions, this becomes $D_{\hat{u}}f(P)=\nabla f(P).\hat{u}$. I don't understand why the existence of first-order partial derivatives by itself is not a sufficient condition for this; why must the function also be differentiable?

Answer 1

There's a good reason the "increment theorem" does not appear as a definition in (good?) mathematics textbooks. The result is false. Although a function satisfying this condition will be differentiable by the usual, correct definition (your "computational check for differentiability"), the converse is in fact false.

What goes wrong is that a differentiable function, say at $(0,0)$, can fail to have a Taylor expansion. Take $g(x,y) = (x^2+y^2)^{\tfrac12+\alpha}$ for $0<\alpha<1/2$. Then the partial derivatives of $g$ at the origin are both $0$, but try to write $$g(x,y) = \epsilon x + \eta y$$ for $x,y\ne 0$. Recall that $(1+u)^\beta = 1+\beta u + \dots$, and so \begin{align*} (x^2+y^2)^{\tfrac 12+\alpha} &= |x|(x^2){}^\alpha\left(1+(\frac yx)^2\right){}^{\tfrac12+\alpha} = |x|(x^2){}^\alpha\left(1+(\tfrac12+\alpha)\frac{y^2}{x^2} + \cdots\right)\\ &= |x|\left((x^2){}^\alpha + (\tfrac12+\alpha) x^{2(\alpha-1)}y^2 + \cdots\right). \end{align*} No matter how you try to configure this as $\epsilon x + \eta y$, you are going to find that one (or both) of $\epsilon$ and $\eta$ will be unbounded along certain paths as $x,y\to 0$. For example, if we take the quantity in parentheses to be $\epsilon$ and $\eta=0$, consider $\epsilon$ along the curve $y=|x|^\alpha$.

Note, by the way, that $g$ is differentiable at $0$, with derivative $0$. By the "computational check" (i.e., the actual definition), we verify that, indeed, $$\lim_{x,y\to 0}\frac{g(x,y)}{\sqrt{x^2+y^2}} = \lim_{x,y\to 0} (x^2+y^2)^\alpha = 0.$$

With regard to the other points of your post, you can check any mathematics text on multivariable analysis (not just a typical engineering calculus book). For your convenience, check out my YouTube lectures — linked in my profile — on partial derivatives, directional derivatives, and differentiability. You will also see the proof that a function with continuous partial derivatives is differentiable, as well as the standard example of a differentiable function whose derivative is not continuous.

EDIT: Not surprisingly, I was incorrect. Note that $$(x^2+y^2)^{\frac12+\alpha} = \frac{(x^2+y^2)^{\frac12+\alpha}}{|x|+|y|}(|x|+|y|).$$ Indeed, $$\lim_{x,y\to 0} \frac{(x^2+y^2)^{\frac12+\alpha}}{|x|+|y|} = 0,$$ since $\frac{(x^2+y^2)^{\frac12+\alpha}}{|x|+|y|} \le (x^2+y^2)^\alpha \to 0$. So set $\epsilon = \text{sgn}(x)\frac{(x^2+y^2)^{\frac12+\alpha}}{|x|+|y|}$ and $\eta = \text{sgn}(y)\frac{(x^2+y^2)^{\frac12+\alpha}}{|x|+|y|}$. What was obvious to me was that we could get upper and lower bounds on the error of the desired form, but, shockingly, I failed to see how to get an equality of the desired form. See, for example, Paul Frost's answer here. Indeed, I should have searched to see that this question was, in fact, a duplicate.