## Relationship Between the Notions of Directional and Total Derivatives (Pt.II)

**Point of Post: **This is a continuation of this post.

What we shall now show that there is an almost converse to the above. Namely, if we impose the not-so-crazy condition that the first partial order derivatives of all types exist on some neighborhood of some point and they are all continuous there then the function is, in fact,totally differentiable there too.

**Theorem: ***Let , where is open, be such that exists everywhere on for and be continuous at . Then, is totally differentiable at .*

**Proof: **It evidently suffices to prove this for mappings by our prior theorem regarding the relationship between the total differentiability of a function and the total differentiability of its coordinate functions. So, let where is open and . Note then that for sufficiently small (so that ) we have that is equal to

(everything just telescopes). But, using logic similar to that in the proof of Clairaut’s theorem we may apply the mean value theorem to find with such that is equal to . Thus, in particular we have that

Thus, we may conclude that

Note though that as we have that and thus by continuity of at for we have that . It thus follows from the above that

Thus, is differentiable at as desired.

*Remark: *This gives an alternate proof (once one knows that total differentiability implies partial differetiability in all directions) that is the matrix of partials–once you combine it with our previous half-uncovered form of the derivative. Also, there is a stronger form of this theorem where one only needs all but one of the partial derivatives to be continuous–but this seems excessive in most cases.

**References:**

1. Spivak, Michael. *Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus.* New York: W.A. Benjamin, 1965. Print.

2. Apostol, Tom M. *Mathematical Analysis*. Reading, MA: Addison-Wesley Pub., 1974. Print.

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