Vector differentiation I have a surface, $S$, defined parametrically as following
$S(u,v)= \langle u^2+v^2,u^2,v^2 \rangle$
its partial derivatives are
$S_u(u,v)= \langle 2u,2u,0 \rangle$
$S_v(u,v)= \langle 2v,0,2v \rangle$
Knowing this, I proceed to create a new surface as following
$S'(u,v)=S(u,v) + T(u,v)*n$
where $S'(u,v)$ is the new surface generated by a point on the surface, $S(u,v)$,
$T(u,v)$, is just a scalar function
and $n$ is the unit surface normal at that given point, that is calculated as following
$n=\dfrac {S_u(u,v) \times S_v(u,v)}{||S_u(u,v) \times S_v(u,v)||}$
My problem is how to obtain the partial derivatives from the new surface
$S'_u(u,v)=S_u(u,v) + T_u(u,v)*n + T(u,v)*n_u$
$S'_v(u,v)=S_v(u,v) + T_v(u,v)*n + T(u,v)*n_v$
I don't know how to calculated the corresponding $n_u$ and $n_v$
The textbook says, $n$ is itself a vector-valued function of $u$ and $v$. The vectors $n_u$ and $n_v$ depend on the curvature of the surface $S$. Normally this curvature is so small that $||n_u||$ and $||n_v||$ are negligible.

This may or may not be correct but lets call $w = S_u(u,v) \times S_v(u,v)$. Then $n = \frac{w}{\vert \vert w \vert \vert}$.
So then calculating out $w$ we get $w = (4 uv, -4 uv, -4uv)$.
Then $n = \left(\frac{u v}{\sqrt{3} \left| u v\right| },-\frac{u v}{\sqrt{3} \left| u v\right|
   },-\frac{u v}{\sqrt{3} \left| u v\right| }\right)$.
Basically $n = \frac{1}{\sqrt{3}} \text{sgn}(u v)(1, -1, -1)$
