Well, intuitively, that's about what I'd expect. The per-vertex normals get interpolated to appear smooth. However, they're only going to be C0 continuous (I think that's the right term). Basically, the rasterized polygons are each going to have smooth normals across the face. Along the edges between faces, things will line up, because the triangle edges are rasterized the same on both sides of the edge. However, as soon as you cross that edge, you're into a new polygon. That polygon is going to have a different slope (in screen space), because the normal is now changing faster or slower (due to the relative sizes of the polygons). Using ddx and ddy, you're simply doing that differentation, so you see the effects of the linear interpolation performed by the rasterizer. [/quote]

But becuase the normals are interpolated, then shouldnt the pixel normal on either side of an edge, when viewed up close, be almost identical? Meaning that the ddx(n) of each pixel quad on either side of an edge be on slightly different?

Well, intuitively, that's about what I'd expect. The per-vertex normals get interpolated to appear smooth. However, they're only going to be C0 continuous (I think that's the right term). Basically, the rasterized polygons are each going to have smooth normals across the face. Along the edges between faces, things will line up, because the triangle edges are rasterized the same on both sides of the edge. However, as soon as you cross that edge, you're into a new polygon. That polygon is going to have a different slope (in screen space), because the normal is now changing faster or slower (due to the relative sizes of the polygons). Using ddx and ddy, you're simply doing that differentation, so you see the effects of the linear interpolation performed by the rasterizer.

But becuase the normals are interpolated, then shouldnt the pixel normal on either side of an edge, when viewed up close, be almost identical? Meaning that the ddx(n) of each pixel quad on either side of an edge be on slightly different?
[/quote]

I'm not sure I follow. Are you just wondering about the effect right along polygon edges, or the faceted look of the whole thing? For the overall effect, it looks like you're drawing a sphere using ~16 segments. So the normals in each segment need to change by a constant amount (360 / 16 degrees). Within each segment, they're going to change at a constant rate in screen space: (360 degrees / 16) / [width in pixels of polygon]. That's highly sensitive to the screen-space coverage of the segments - and you can see this in your images. The segments near the 'front' of the sphere have a much smaller change in the dndx color than those near the edge. As you approach the silhouette of the sphere, the rasterized size of the segments approaches zero very quickly.

For another way of thinking - you're trying to differentiate a function (the sphere normal). It *looks* like it's a smooth/continuous function, but it isn't. Because of the polygons and the rasterizer, it's actually a piecewise-linear approximation of the true result. When you differentiate a piecewise-linear function, you get a step function that is just the constant slope of each linear segment...

Indeed, the rate of change maintains constant between the two vertices. That's because mathematically everything that varies per pixel is canceling each other. If you manage to put some variation to the formula so that elements unique to each pixel remains, you may get smooth results as you wanted.

Try normalizing the normal before dd'ing, that may do the trick:

float4 PS(VS_OUT pIn) : SV_Target
{
float3 n = normalize( pIn.normalW );
float3 dndx = ddx( n);
float3 dndy = ddy( n);

float sn = 100.0f;
return float4( n, 1 );
return float4( sn*dndx, 1 );
return float4( sn*dndy, 1 );
}