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
$\dfrac{\partial T}{\partial q} =\dfrac{ (2 q r^{1/2}+q^{1/2} r)-( q^{1/2} r)(2 \sqrt r+\dfrac{r}{2 \sqrt q})}{(2q\sqrt r+\sqrt qr)^2}$
and $\dfrac{\partial T}{\partial p} =0$
Plug the values $p=2; q=1; r=4$, we get
$\dfrac{\partial T}{\partial q} =\dfrac{ (2 q r^{1/2}+q^{1/2} r)-( q^{1/2} r)(2 \sqrt r+\dfrac{r}{2 \sqrt q})}{(2q\sqrt r+\sqrt qr)^2}=\dfrac{ (-8)}{64}=\dfrac{-1}{8}$;
and
$\dfrac{\partial T}{\partial r} =\dfrac{ (2 q r^{1/2}+q^{1/2} r)(\sqrt q)-(q^{1/2} r)(\dfrac{q}{\sqrt r}+\sqrt q)}{(2q\sqrt r+\sqrt qr)^2}$;
Plug the values $p=2; q=1; r=4$.
$\dfrac{\partial T}{\partial r} =\dfrac{ (2 q r^{1/2}+q^{1/2} r)(\sqrt q)-(q^{1/2} r)(\dfrac{q}{\sqrt r}+\sqrt q)}{(2q\sqrt r+\sqrt qr)^2}=\dfrac{2}{64}=\dfrac{1}{32}$
