Answer
$p_t=\frac{2t^3}{\sqrt{t^4+u^2\cos v}}$, $p_u=\frac{u\cos v}{\sqrt{t^4+u^2\cos v}}$, $p_v=\frac{-u^2\sin v}{2\sqrt{t^4+u^2\cos v}}$.
Work Step by Step
$p=\sqrt{t^4+u^2\cos v}$
In order to find $p_t$ we treat $u$ and $v$ as constants and differentiate with respect to $t$.
$p_t=\frac{2t^3}{\sqrt{t^4+u^2\cos v}}$
Analogously:
$p_u=\frac{u\cos v}{\sqrt{t^4+u^2\cos v}}$
$p_v=\frac{-u^2\sin v}{2\sqrt{t^4+u^2\cos v}}$
