In real and complex analysis, one has a powerful tool, namely the Taylor expansion, which expands an analytic function into a power series. In algebra, one can define the derivative of a polynomial aswell; for , being a unitary ring, define . This satisfies the same rules as the usual derivative, for example linearity and the product rule . One can also define recursively by , for . Unfortunately, one looses certain properties; for example, if is of finite characteristic , the polynomial satisfies , but is not constant as (assuming is not the zero ring). In particular, the Identity Theorem does not work. This example also shows one problem with a possible Taylor expansion: for that, one needs to compute for ; but in , whence has no inverse in ! Hence, a Taylor expansion in the classical sense cannot be defined. A “fix” for this problem is offered by Hasse derivatives: they are defined to make both the Identity Theorem and Taylor expansions work again.
Let and . Define
The function is called the th Hasse derivative.
The Hasse derivative shares several properties with the usual derivative, but not all of them; for example, in general. But we have the following properties:
Let and , and .
 We have that is linear, i.e. and .
 We have ; in particular, .
 We have .

(Leibniz Rule) We have
more generally, for , we have
where the sum goes over all such tuples .

(Faà di Bruno's Formula) We have
where the sum goes over all tuples with ; here, is a multinomial coefficient having the value

(Taylor Formula) We have
 (Identity Theorem) If we have for all , then .
For that reason, one can define , so that we can write Taylor's formula in a more tempting form as
which almost equals the classical form.
Note that the Leibniz rule, Faà di Bruno's formula and Taylor's formula take simpler forms than their classical counterparts; this is due to the fact that the additional factorial terms or binomial coefficients already hide in the definition of the Hasse derivative.
 This follows from the definition of .

Write ; then

By 1., it is suffices to show this for , (for smaller , both sides will be zero). We have
and
But since
these terms are equal.

Note that if we fix , we get an linear function , . Hence, it suffices to show this for arbitrary and , . By the same argument, for , we get an linbear function , ; therefore, it suffices to consider , . But now,
Hence, it suffices to show . By reorganizing the binomial coefficients, one transforms this into the equality
which is Vandermonde's Identity and, hence, true.
The more general equation is shown by induction on . For , we have . Now, assume that the equation is true for all for one . Then, for any ,
by the Leibniz rule and by the induction hypothesis; here, the second sum goes over all such tuples . But this equals , what we had to show.

Again, by 1., it suffices to show this for , . Now, by the second part of 4.,
where the sum goes over all such . The formula we want is now obtained by sorting the summands by the different powers of appearing, .
Consider the map
there , . Now, if satisfies , then and . Now, for a fixed with and , the number equals the multinomial coefficient
whence we get that the above formula for equals
where the sum goes over all tuples with and .
Now note that and . Therefore, we get
If one now replaces by , we obtain the formula from the statement.

By 1., it suffices to show this for , . Now
by the Binomial Theorem, what we had to show.
 This follows directly from 6.
Comments.