In ideal gas $\partial_V(PdV)_T=P$ First $\partial_V(PdV)_T=\partial_V(PV)_T=P$
However, if write it by chain rule and $P=nRT/V$
$\partial_V(PdV)_T=P\partial_V(dV)_T+\partial_V(P)_TdV=P\cdot 1+\partial_V(P)_TdV=P+P=2P$.
What's wrong with the second expression? Or why does $\partial_V(dV)_T$ or $\partial_V(P)_T$ equal to $0$.
 A: So basically you've made two errors in your above question. The first is in the first expression. Namely: it is ${\bf not\, true}$
that 
$$
\left[\frac{\partial (PdV)}{\partial V}\right]_T = \left[\frac{\partial (PV)}{\partial V}\right]_T = P\,.
$$
In the second expression you try to apply the chain rule to write
$$
\left[\frac{\partial (PdV)}{\partial V}\right]_T = P\left(\frac{\partial (dV)}{\partial V}\right)_T + \left(\frac{\partial P}{\partial V}\right)_T dV\,,
$$
which does not work because taking the partial derivative of an infinitesimal volume with respect to the volume doesn't make sense.
In the first expression, to see why 
$$
\left[\frac{\partial (PV)}{\partial V}\right]_T = P
$$
is ${\bf not\, true}$ consider what is written below.
For an ideal gas
$$
PV = nRT\,.
$$
So 
$$
\left[\frac{\partial (PV)}{\partial V}\right]_T =\left[\frac{\partial (nRT)}{\partial V}\right]_T\,.
$$
Since $T$ is held constant and assuming the number of moles $n$ is constant
$$
\left[\frac{\partial (PV)}{\partial V}\right]_T =0\,.
$$
But
$$
\left[\frac{\partial (PV)}{\partial V}\right]_T = P\left(\frac{\partial V}{\partial V}\right)_T + V\left(\frac{\partial P}{\partial V}\right)_T\,.
$$
Thus
$$
P(1) + V\left(\frac{\partial P}{\partial V}\right)_T = 0\,.
$$
Therefore
$$
P = - V\left(\frac{\partial P}{\partial V}\right)_T\,.\quad\,(*)
$$
Since an ideal gas is being considered, the pressure can be expressed as 
$$
P = \frac{nRT}{V}\,.
$$
From this it follows that
$$
\left(\frac{\partial P}{\partial V}\right)_T = -\frac{nRT}{V^2}\,.
$$
Therefore
$$
V\left(\frac{\partial P}{\partial V}\right)_T = -\frac{nRT}{V}\,.
$$
Substituting this into the right hand side of the equation marked with a $(*)$ it is found that
$$
P = \frac{nRT}{V}\,.
$$
This is a result that can be obtained from the ideal gas law directly and which we knew already.
