Differentiability of a scalar function of two variables --- problems with Stewart’s definition.

There are functions for which all directional derivatives exist and are still not differentiable. A web search will turn up several examples such as this one, in which not only do they all exist but are equal. It fails to be differentiable because all paths need to be considered, not just the straight lines. Requiring continuity as well likely does the trick, but to me that seems like using a bigger and bigger hammer to deal with each issue as it arises.

I think the awkwardness in the text’s definition comes from treating $x$ and $y$ separately. If you formulate the definition of differentiability in terms of vectors, it doesn’t seem all that awkward to me: A function $f:\mathbb R^m\to\mathbb R^n$ is differentiable at $\mathbf v\in\mathbb R^m$ if there exists a linear map $L_{\mathbf v}:\mathbb R^m\to\mathbb R^n$ such that $f(\mathbf v+\mathbf h) = f(\mathbf v)+L_{\mathbf v}(\mathbf h)+o(\mathbf h)$. If such a map exists, it isn’t hard to show that it is unique. This linear map is the differential of $f$ at $\mathbf v$, often denoted by $\mathrm df_{\mathbf v}$ or simply $\mathrm df$. This captures the essential notion that a derivative/differential is the best linear approximation to the the change in function’s value near a given point.

Note that when $f:\mathbb R\to\mathbb R$ this definition reduces to that of the derivative from elementary calculus (in one dimension, a linear map is just multiplication by a scalar). In addition, gradient, directional derivative, &c can all be defined in terms of this differential.

Define $f$ to be $0$ on the axes, and $f(x,y) = \sqrt { x^2+y^2}$ elsewhere. Then $f$ is continuous at $(0,0),$ and $\partial f/\partial x (0,0) = 0 = \partial f/\partial y (0,0).$ Suppose $f$ is differentiable at $(0,0).$ Verify that the definition would then yield $f(x,x)=o(x)$ as $x\to 0.$ But $f(x,x) = \sqrt 2 |x|,$ contradiction, showing $f$ is not differentiable at $(0,0).$