bloodhound Posted December 2, 2004 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.
MandrakeRoot Posted December 3, 2004 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
Dave Posted December 4, 2004 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.
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