Partial derivative of $x$ w.r.t. $\bar z$? My book does this :

$x=\frac {(z+\bar z)}2$. Then $\frac {\partial x}{\partial \bar z}=\frac 12$. 

But this doesn't make sense to me as $\bar z$ is a function of $z$. 
Would we say that if $y=x+x^2$ then $\frac {\partial y}{\partial x^2}=1$?
 A: The operator $\partial \over \partial \overline{z}$ has a very precise definition, unlike $\partial \over \partial x^2$ that you just made up to make a point.
That definition is $\frac 12({\partial \over \partial x}+i{\partial \over \partial y})$ where $\partial \over \partial x$ and $\partial \over \partial y$ work the way you imagine.
Therefore you can see that $\partial \over \partial \overline{z}$ is linear, and satisfies ${\partial \over \partial \overline{z}}(z)=0$ and ${\partial \over \partial \overline{z}}(\overline{z})=1$.
A: We are talking about Wirtinger derivatives here:$$\frac{\partial f}{\partial z}=\frac12\left(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right)\text{ and }\frac{\partial f}{\partial\overline z}=\frac12\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right).$$
A: If we consider 
$$x=x(z,\bar z)=\frac{z+\bar z}{2}$$
then by chain rule
$$\frac{\partial x}{\partial z}=\frac12\qquad\frac{\partial x}{\partial \bar z}=\frac12$$
which simply means that
$$dx=\frac{\partial x}{\partial z}dz+\frac{\partial x}{\partial \bar z}d\bar z=\frac12(dz+d\bar z)$$
