bloodhound Posted December 2, 2004 Share Posted December 2, 2004 Let [math]f\colon \mathbb{R}^{2}\to\mathbb{R}[/math] given by [math]f(x,y)=\frac{xy^3}{x^2+y^6}[/math] if [math](x,y)\ne (0,0)[/math] and 0 if (x,y)=(0,0) Question asks me to use the definition or partial derivatives to show that f_x and f_y both exist at (0,0). and at any other point (x,y) i can do the first part, but for the second part i cannot simplify the expression for the limit. Link to comment Share on other sites More sharing options...

MandrakeRoot Posted December 3, 2004 Share Posted December 3, 2004 Can't you use the binomial formula or something like that ? If you can do it at (0,0), i think you are surely pretty close to showing it for every couple (x,y) Mandrake Link to comment Share on other sites More sharing options...

Dave Posted December 4, 2004 Share Posted December 4, 2004 The expression for the limit does simplify: first simplify the expression for f(x+h,y) - f(x,y) by making it into a single fraction, then after that you should find that all the terms on the top of the fraction that do not have a h in them will cancel, and then the other h's will cancel with the one on the bottom. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now