Determining tangent space Let $g(x,y)=\sqrt{y-x}$ defined on $\{(x,y):y>x\}$ and let manifold $M=\{(x,y,z):z=g(x,y)\}$.
I want to determine the tangent space in $(x,y,z)=(2,6,*)$
I usually solved this sort of tasks using that the tangent space in $(x,y,z) \in M$ is $\{v \in \mathbb{R}^3:\nabla gf(x,y,z)(v)=0\}$. But I don't know how to start here.
I would appreciate it if someone could show me how to this.
Thank you!
 A: (1) The manifold $M$ is the graph of the function $g$. Note that this manifold (surface) is parametrized by
$$ (x,y)\mapsto \left(x,y,g(x,y)\right) =\left(x,y,\sqrt{y-x}\right),$$
for $y>x$. Deriving this map w.r.t. $x$ and $y$ gives two linearly independent tangent vectors
$$ \begin{align*}
\left(1,0,\frac{\partial g}{\partial x}(x,y)\right) &= \left(1,0,\frac{-1}{2\sqrt{y-x}}\right) \\
\left(0,1,\frac{\partial g}{\partial y}(x,y)\right) &= \left(0,1,\frac{1}{2\sqrt{y-x}}\right).
\end{align*}
$$
So the tangent space at $(x,y,z)=(2,6,2)$ is the plane with parameter equation.
$$
 (2,6,2) + r \left(1,0,-\frac{1}{4}\right) + s \left(0,1,\frac{1}{4}\right), \quad \text{$r,s\in\mathbb{R}$.}
$$
This is probably the easiest way.
(2) Your attempt is also possible, but there is a serious typo in your question. 
Note that $M$ is given by $f(x,y,z)=0$ where $f(x,y,z)=z-g(x,y)$. The gradient $\nabla f(x,y,z)$ gives a vector that is normal to the tangent space at $(x,y,z)$. Here 
$$\nabla f(x,y,z) = \left(\frac{1}{2\sqrt{y-x}},\frac{-1}{2\sqrt{y-x}},1\right),$$
so $\nabla f(2,6,2)=(1/4,-1/4,1)$.
Now you need to find two linearly independent vectors that are normal to this vector, and you will get the tangent space.
