## Uniqueness of the Differential

Okay, for the moment let’s pick an orthonormal basis for our vector space . This gives us coordinates on the Euclidean space of points. It also gives us the dual basis of the dual space . This lets us write any linear functional as a unique linear combination . The component measures how much weight we give to the distance a vector extends in the direction.

Now if we look at a particular point we can put it into our differential and leave the second (vector) slot blank: . We will also write this simply as , and apply it to a vector by setting the vector just to its right: . Now is a linear functional, and we can regard as a function from our space of points to the dual of the space of displacements. We can thus write it out uniquely in components , where each is a function of the point , but not of the displacement .

We want to analyze these components. I assert that these are just the partial derivatives in terms of the orthonormal basis we’ve chosen: . In particular, I’m asserting that if the differential exists, then the partial derivatives exist as well.

By the definition of the differential, for every there is a so that if , then

Now we can write out in components

Next for a specific index we can pick for some value . Then , , and for all the other indices . Putting all these and the component representation of into the definition of the differential we find

Dividing through by we find

And this is exactly what we need to find that exists and equals .

Therefore if the function has a differential at the point , then it has all partial derivatives there, and these uniquely determine the differential at that point.