# Partial derivatives of polynomial in two variables

Let $k \in \mathbb N$, $a_{ij} \in \mathbb R$ for $i,j \in \mathbb N$, $i+j \le k$. A function $f:\mathbb R^2 \to \mathbb R$

$$f(x,y) = \sum_{i+j \le k} a_{ij}x^iy^j$$

is called polynomial of degree $k$ in $x,y$.

I have to show that the partial derivatives of polynomials of degree $k \ge 1$ exist, and the partial derivatives $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$ are polynomials of degree $k-1$.

I start showing that for a polynomial of degree $1$ its partial derivatives $D_1f(x)$ and $D_1f(y)$exists and in the next step that the $D_if(x)$ and $D_if(y)$ derivative exists, but I dont know how to deal with the matrix and the indices of the sum in the formula, how can I take a take a partial derivative of it? Dealing with the matrix and other variable as a constant and use normal derivative rules for the variable in which I want to take the partial derivative?

Any help is appreciated!!!

• I believe that $y$ should be raised to the power $j$ in your formula, as opposed to $i$? – Foo Barrigno Apr 24 '14 at 17:28
• yes, I edited it, thanks for the suggestion – sj134 Apr 24 '14 at 17:37

There is at least one caveat to your problem that needs to be stated. At least one $a_{ij}$ such that $i+j=k$ must be non-zero. Otherwise your polynomial is of degree less than $k$. And in order for the partial derivative w.r.t. $x$ to be a polynomial of degree $k-1$ one of the $a_{ij}$ must have $i\ge 1$. Similarly for $y$.
Otherwise, show that if $i+j=k$, the partial derivative of $a_{ij}x^iy^j$ for both $x$ and $y$ is of total degree $i+j-1=k-1$.
To actually perform the partial derivatives, note that: $$\frac{\partial f}{\partial x} \sum_{i+j \le k} a_{ij}x^iy^j = \sum_{i+j \le k} i*a_{ij}x^{i-1}y^j$$ and vice versa for the partial derivative with respect to y. Just note that in cases where there is no $x$ the partial derivative with respect to $x$ is 0.