Finding the equation of the tangent plane to the surface $z=x^2+ y^2$ at the point $(1,2)$ Just wanted to ask a simple question I've forgotten how to solve (lost my mind completely).

Find the equation of the tangent plane to the surface $z=x^2 + y^2$
  at $(1,2)$.

The fact it has $3$ variables is what is putting me off. I don't know whether to differentiate with respect to $x$ or $y$.
 A: Let $M$ denote the surface described by your equation. One way of writing down that plane is to find a point in the plane (you can take the one you are given, $p=(1,2, 5)$) and calculate two tangent vector $v, w$ and then let the tangent plane to the surface M in $p$
$$T_pM = \{q\in \mathbb{R}^3: q = p + tv +sw; s, t \in \mathbb{R}\}$$
It remains to find $v, w$. These you can take as the derivative in $t=0$ of a curve in $M$ through $p$ like, in your case, $(c(t), 2, c^2(t)+4)^T$ and $(1, \bar{c}(t), \bar{c}^2(t)+1)^T$ such that $c(0)=1, \bar{c(0)}= 2$. You may take $c(t) = 1 +t $ and $\bar{c}(t)=2+t$.
This results in, e.g., $v=(1, 0, 1)^T$ and $w=(0,1,4)^T$
A: This plane has normal $(-\dfrac{\partial z}{\partial x}(1,2),-\dfrac{\partial z}{\partial y}(1,2),1) = (-2,-4,1)$. 
Then the plane has the form
\begin{equation}
-2x -4y + z + d = 0.
\end{equation}
And the plane pass by (1,2,5), hence the plane is
\begin{equation}
-2x -4y + z + 1 = 0.
\end{equation}
A: Let $z:\mathbb{R}^2\rightarrow \mathbb{R}$ be differentiable at $(x_{0},y_{0})$. The plane in $\mathbb{R}^3$ defined by the equation
$$T_{p}=z(x_{0},y_{0})+\left[\frac{\partial z}{\partial x}(x_{0},y_{0})\right](x-x_{0})+\left[\frac{\partial z}{\partial y}(x_{0},y_{0})\right](y-y_{0})$$ is called the tangent plane of the graph of $z$ at the point $(x_{0}, y_{0})$. 
This is the definition, now we have to find the tangent plane for $z=x^2+y^2$ at the $(1,2)$. 
$$z(1,2)=5;$$
$$\left[\frac{\partial z}{\partial x}(1,2)\right](x-1)=2(x-1)=2x-2 ;$$
$$\left[\frac{\partial z}{\partial y}(1,2)\right](y-2)=4(y-2)=4y-8 .$$
So, the equation for tangent plan at point $(1,2)$ is : 
$$T_{p}=5+2x-2+4y-8=2x+4y-5. $$
