Derivative of Normal Vector Field

Question

This is an example from Do Carmo (Example 4, page 139). Consider the saddle point $p=(0, 0, 0)$ of the hyperbolic paraboloid $z=y^2-x^2$ with parameterization $\mathbf x(u, v)=(u, v, v^2-u^2)$. It is not hard to find the unit normal vector field as $$N(u, v)=\mathbf x_u\times\mathbf x_v=\left(\frac{u}{\sqrt{u^2+v^2+1/4}}, \frac{-v}{\sqrt{u^2+v^2+1/4}}, \frac{1}{2\sqrt{u^2+v^2+1/4}}\right).$$ Then it states that the derivative of $N$ at $p$ is $\boxed{N'(0)=\left(2u'(0), -2v'(0), 0\right)}$. How to get this final result, especially, the "2" and "-2" in it, please? Thank you!

Answer 1

For the first entry, by the quotient rule:

$(\frac{u}{\sqrt{u^2+v^2+1/4}})'=\frac{(\sqrt{u^2+v^2+1/4})u'-u(\sqrt{u^2+v^2+1/4})'}{u^2+v^2+1/4}$

Evaluating this at zero we get:

$\frac{(1/2)u'(0)}{(1/4)}=2u'(0).$

Now you can do the other entry.