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A non-differentiable function with partial derivatives everywhere.

Define f by


Then tex2html_wrap_inline489 and tex2html_wrap_inline491 certainly exist at every point except (0,0), and may be calculated by the quotient rule. They also exist at (0,0), although you need to use the definition of partial derivative to find them:


and tex2html_wrap_inline499 may be found similarly to equal 0. However f is not differentiable at (0,0). In fact, f is not even continuous at (0,0)! To see this, approach the origin along the line y=x. Since f(x,x)=1/2 for all tex2html_wrap_inline205 , the limit along this line is 1/2. But along the line x=0, f(0,y)=0 for tex2html_wrap_inline521 , so along this line f approaches 1/2. Therefore the two dimensional limit doesn't exist. Here's a picture of f:


Although the partials of this function exist at every point, they can't be continuous everywhere, since there is a theorem telling us that functions with partial derivatives which are continuous in an open set must be differentiable in that set.

Tom Vogel
Mon May 5 12:53:33 CDT 1997