How to compute dPdu, dPdv for triangles

For triangles, as well as other planar shapes, there's an affine mapping from parametric space to object space, said M.

In parametric space, we know the parametric coordiantes for 3 triangle corners are (u0, v0), (u1, v1), (u2, v2) respectively.

While in object space, the 3D coordinates for 3 triangle corners are P0, P1, P2.

So with the mapping M, we can transform the difference vectors (du1, dv1) = (u1 - u0, v1 - v0), and (du2, dv2) = (u2 - u0, v2 - v0) to the difference vectors in object space, said dP1 = P1 - P0, dP2 = P2 - P0, dP1, dP2 are 3D vectors.

The equation looks like:

(dP1)    (du1 dv1)

|     | = |           | * M

(dP2)    (du2 dv2)

Assume M = (K1, K2), K1, K2 are 3D vectors, the equation can be rewritten with:

(dP1)    (du1 dv1)   (K1)

|     | = |           | * |   |       (1)

(dP2)    (du2 dv2)   (K2)

Solve K1, K2, we get:

(K1)         (du1 dv1)   (dP1)

|    | = INV|           | * |    |

(K2)         (du2 dv2)   (dP2)

Now we need to understand what are K1, K2, using equation (1) we have:

dPdu * du = (du 0) * (K1) = K1 * du, thus we know dPdu = K1

                               (K2)

Similarly,

dPdv * dv = (0 dv) * (K1) = K1 * dv, thus we know dPdv = K2

                               (K2)

Finally we got:

(dPdu)         (du1 dv1)   (dP1)

|       | = INV|           | * |    |

(dPdv)         (du2 dv2)   (dP2)

Which satisfies:

P(u + du, v) = P(u, v) + dPdu * du

P(u, v + dv) = P(u, v) + dPdv * dv