Are these definitions of the total derivative equivalent?

Question

The total derivative of a function $F:\mathbb{R}^n \to \mathbb{R}^m$ is defined in this way:

If a linear map $L:\mathbb{R}^n\to\mathbb{R}^m$ exists such that \begin{equation} \tag{*} \lim_{\boldsymbol{h}\to\boldsymbol{0}} \frac{||F(\boldsymbol{x}+\boldsymbol{h})-F(\boldsymbol{x}) - L\boldsymbol{h}||}{||\boldsymbol{h}||} = 0 \end{equation} then we say $L$ is the total derivative of $F$ at $\boldsymbol{x}$ and we denote it by $D_x F$.

This is analogous to a possible definition of the derivative in the single-variable case for a function $f:\mathbb{R}\to\mathbb{R}$:

If a number $l\in\mathbb{R}$ exists such that $$ \lim_{h\to 0} \frac{f(x+h)-f(x)-lh}{h} = 0 $$ then we say $l$ is the derivative of $f$ at $x$ and we denote it by $f'(x)$.

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However, a more standard direct definition of the derivative in the single variable case is usually used:

$$ f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} $$ Where the definition holds if the limit on the right exists.

Is a similar direct definition possible for the multivariate total derivative?

\begin{equation} \tag{**} D_x F \: \boldsymbol{v}= \lim_{h\to 0} \frac{F(\boldsymbol{x}+h\boldsymbol{v}) - F(\boldsymbol{x})}{h} \end{equation} Where the definition holds if the limit on the right exists.

Are the definitions $(*)$ and $(**)$ of $D_xF$ equivalent in all cases?

Answer 1

No, because $(**)$ is the definition of the directional derivative, not the derivative. You have examples of functions that have directional derivatives in all directions, i.e. $D_x F \: \pmb{v}$ exists for all directions $\pmb {v}$, but the function is not differentiable.
Classical example is $f(x,y)=\frac{x^3}{x^2+y^2}$ considered at the origin $(0,0)$. In polar coordinates the function is ${f}(r,\theta)=r \cos^3(\theta)$. Thus $\frac{\partial f}{\partial r}= cos^3(\theta)$, so directional derivative at $(0,0)$ exists in all directions, but the function is non-differentiable since there is no linear map $L$ that satisfies $(*)$. You can see it by noting that $(*)$ would imply that all directional derivatives are linear combinations of the partial derivatives $(\frac{\partial f }{\partial x},\frac{\partial f }{\partial y})$. Essentially, being differentiable at a point means you can place a tangent plane at that point. You can have directional derivatives is all directions but not have such a tangent plane - such as the above example illustrates.

Answer 2

No, these are not equivalent. The reason is that $(*)$ allows for more than just straight paths along which $\bf h$ can approach 0, while $(**)$ allows for only straight paths, since $h\bf v$ is always on the same straight line.

For instance, consider the following function:

$$ f:\mathbb R^2\to\mathbb R, \\ f(x,y)=\begin{cases} 0&\textrm{if }y=x^2\\ x&\textrm{else} \end{cases} $$

If we graph this function, it's basically a constantly sloping plane, except on the parabola $y=x^2$, where it is 0. According to definition $(*)$, this is not differentiable at $(0,0)$, because the only possible differential is $(1,0)$ (by calculating partial derivatives), but $\lim_{h\to0}\frac{f(h,h^2)-f(0,0)-(1,0)\cdot(h,h^2)}{h}=1$. However, according to definition $(**)$, this is differentiable, since the corresponding limits exist.

Your definition $(**)$ hints at the directional derivative: The limit you're calculating is exactly the directional derivative of $f$ along the direction $v$, and it is true that if $f$ is totally differentiable, then this directional derivative is equal to $\mathrm D_xfv$. But the definition of total differentiability is much more restrictive, requiring some sense of smoothness along all, possibly even curved, directions along which a point on the graph can be approached.