3D - derivative of a point's function, is it the tangent? If I have (for instance) this formula which associates a $(x,y,z)$ point $p$ to each $u,v$ couple (on a 2D surface in 3D):
$p=f(u,v)=(u^2+v^2+4,2uv,u^2−v^2) $
and I calculate the $\frac{\partial p}{\partial u}$, what do I get? The answer should be "a vector tangent to the point $p$" but I can't understand why. Shouldn't I obtain another point?

Take a fixed location where $(u,v) = (u_0,v_0)$. Think about the mapping $u \mapsto f(u,v_0)$. This is a curve lying on your surface, which is formed by allowing $u$ to vary while $v$ is held fixed. In fact, in my business, we would say that this is an "isoparametric curve" on the surface. By definition, $\frac{\partial f}{\partial u}(u_0)$ is the first derivative vector of this curve at $u= u_0$. In other words, it's the "tangent" vector of this curve.
