Answer
$\dfrac{\partial T}{\partial p} =0$; $\dfrac{\partial T}{\partial q} =\dfrac{-1}{8}$;
and $\dfrac{\partial T}{\partial r} =\dfrac{1}{32}$
Work Step by Step
$\dfrac{\partial T}{\partial p} =0\\\dfrac{\partial T}{\partial q} =\dfrac{ (2 q \sqrt {r}+\sqrt q r)-(\sqrt qr)(2 \sqrt r+\dfrac{r}{2 \sqrt q})}{(2q\sqrt r+\sqrt qr)^2}$;
$\dfrac{\partial T}{\partial q} =\dfrac{ -8}{64}=\dfrac{-1}{8}$; (When $p=2; q=1; r=4$)
$\dfrac{\partial T}{\partial r} =\dfrac{ (2 q \sqrt {r}+\sqrt q r)(\sqrt q)-(\sqrt qr)(\dfrac{q}{\sqrt r}+\sqrt q)}{(2q\sqrt r+\sqrt qr)^2}$; (When $p=2; q=1; r=4$)
Thus, we have $\dfrac{\partial T}{\partial r} =\dfrac{2}{64}=\dfrac{1}{32}$
