Question

Consider a function $f(x_1,...,x_n)$of $n$variables. How many different partial derivatives of order $r$does $f$possess?

Short answer

The different partial derivatives of order $r$that $f$possess arelocalid="1648374000028" $\frac{n+r-1!}{(r-1)!n!}$.

Step-by-step solution

Step 1. State the proposition.

The proposition used here is:

There are $\left(\begin{array}{c}n+r-1\\ r-1\end{array}\right)$distinctive non negative integer valued vectors localid="1648373873954" $\left(x_1,x_2,.......,x_n\right)$satisfying the equation$x_1+x_2+......+x_n=n$.

Step 1. Find the number of partial derivatives.

The given function $f(x_1,...,x_n)$consists of $n$variables.

We have to find the partial derivatives of order $r$possessed by $f$.

This means we have to non negative integer valued vectors of $\left(x_1,x_2,.......,x_n\right)$

Therefore, from the above proposition, the different partial derivatives of order $r$possessed by $f$ is given by

$\frac{n+r-1!}{(r-1)!n!}$.