# Proving differentiability

• Mar 21st 2009, 05:14 PM
Pinkk
Proving differentiability
Determine whether or not $\displaystyle f(x,y)=x^{2}+3xy$ is differentiable.

Now, from what I understand, in order to show that a function of two variables is differentiable at some point $\displaystyle (a,b)$, I have to show that $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}$ are continuous at $\displaystyle (a,b)$ and that $\displaystyle f(x,y)$ itself is continuous at $\displaystyle (a,b)$. I know that the partials are continuous for all $\displaystyle (x,y)$, but how do I show that $\displaystyle f(x,y)$ is continuous? (The problem does not ask if the function is differentiable at a specific point $\displaystyle (a,b)$.) Thanks in advance.

Same question with the equation $\displaystyle f(x,y)=xy^{2}$.
• Mar 21st 2009, 10:23 PM
matheagle
I think all they want is that, since $\displaystyle x^2$, $\displaystyle x$ and $\displaystyle y$ are all differentiable, then their product and sum is differentiable.
This is not true in regards to division, like $\displaystyle 1/x$.
• Mar 21st 2009, 10:35 PM
Pinkk
So since there are no points where either of those equations are undefined, the equations are continuous?
• Mar 21st 2009, 10:43 PM
matheagle
All three (x,y, x squared)are continuous everywhere
1/x is continuous everywhere except at zero, where it's not even defined.
• Mar 22nd 2009, 04:35 AM
HallsofIvy
Quote:

Originally Posted by Pinkk
So since there are no points where either of those equations are undefined, the equations are continuous?

All polynomials have the property that where they are defined they are continuous (and, in fact, infinitely differentiable). That is, of course, not true for all functions.
• Mar 22nd 2009, 06:21 AM
Pinkk
Ah, okay. I wanted to make sure that was true for functions of two variables in three dimensions.