# What is this notion of differentiability?

The standard definition for the derivative of a function $$f: R^n \to R^m$$ at a point $$x_0$$ is the unique linear map, $$f'(x_0): R^n \to R^m$$, such that $$\lim_{x \to x_o}\frac{\|f(x) - f(x_0) - f'(x_0)(x - x_0)\|}{\|x - x_0\|} = 0$$

On the other hand, the brain-dead generalization of the single-variable definition might be something like this: $$f'_{stupid}(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{\|x - x_0\|}$$

which seems to say something like "no matter what direction I take to approach $$x_0$$, the "$$m$$-d slope'' of $$f$$ is the same and is equal to $$f'_{stupid}(x_0$$)." Or, on other words, that all directional derivatives are equal to $$f'_{stupid}(x_0)$$.

Can anyone shed some light on any relations between the two?

• It is worth noting that the first definition actually bakes in the concept of a "direction" being applied. Your second definition almost surely doesnt exist unless the function has 0 slope, because any slanted surface will have a different value of the derivative depending on the direction you approach from (notably the denominator is always positive, unlike the "h" we see in the usual derivative). Commented Aug 8 at 20:53
• Yeah I’m pretty sure it only exists for constant functions and is equal to zero haha Commented Aug 8 at 20:53
• I would think $f(x) = 1 + \lVert x \rVert$ has $f'_{stupid}(0) = 1$, though of course 0 would then be the only point where $f'_{stupid}(x)$ exists. Commented Aug 8 at 22:57
• It might be worth noting that in your brain-dead-definition for $m=1$, if you replace the lim with a limsup, you more or less get what is called a "local slope" (up to a few technicalities). In $R^n$ this is of course less useful, but the point is that the definition still works in metric spaces, as a replacement of $|\nabla f|$.
– mlk
Commented Aug 9 at 11:23

The thing to keep in mind is that in order for a multidimensional limit $$\lim_{x \to x_0}$$ to exist, the value of the limit must be the same no matter how $$x$$ approaches $$x_0$$. This even holds in the 1-dimensional case of course: we require the left and right limits to be equal, in order for the limit to exist.
So, let's suppose that the first (correct) definition holds, and consider the linear transformation $$f'(x_0) : R^n \to R^m$$. Suppose that $$f'(x_0)$$ is not the zero transformation, in other words there exists a unit vector $$u \in R^n$$ such that the image vector $$w = f'(x_0)(u)$$ is not the zero vector. Then the limit that defines $$f'_{stupid}(x_0)$$ does not exist. To prove this, first let $$x$$ approach $$x_0$$ along the ray $$x_0 + t u$$ for $$t > 0$$ you get a certain nonzero value which is $$\lim_{t \to 0^+} \frac{f(x_0+tu) - f(x_0)}{\| t u \|} = f'(x_0)(u) = w$$ whereas if you let $$x$$ approach $$x_0$$ along the ray $$x_0 + t \vec u$$ for $$t < 0$$ then you get the opposite value $$\lim_{t \to 0^-} \frac{f(x_0+tu) - f(x_0)}{\| t u \|} = f'(x_0)(-u) = -w$$ The exceptional case is that $$f'(x_0)$$ is the zero transformation, in which case all directional deriatives are zero, the limit in the brain dead definition exists, and its value is zero.