The picture on the left shows a graph of f(x,y) = (x2 + y2) sin (x2 + y2)-1/2 The red curve shows the cross section x=0, while the green curve highlights the cross section y=0. Click and drag on the picture to rotate it; type "F" after clicking on the picture to view the cross sections without the surrounding surface.1 This function is differentiable at (0,0), and both fx(0,0) and fy(0,0) equal 0. These partials exist for essentially the same reason that x2sin(1/x) is differentiable in Calculus 1. In fact, it's exactly the same reason: If y=0, then f(x,0)=x2sin(1/|x|), If x=0, then f(0,y)=y2sin(1/|y|) Most calculus books include a theorem saying that if the partials are continuous at and around a point, a function is differentiable. This example shows that a function can still be differentiable even if its partials are not continuous. (You can verify this on your own, but both partials include a term that oscillates wildly as (x,y) approaches the origin.) 1 Why F? The applet on this page uses F to toggle the display of polygon Faces, which are the computer graphics objects which make up the displayed surface.