Answer
$v_{rr}=0$
$v_{rs}=-sin(s+2t)$
$v_{rt}=-2sin(s+2t)$
$v_{sr}=-sin(s+2t)$
$v_{ss}=-rcos(s+2t)$
$v_{st}=-2rcos(s+2t)$
$v_{tr}=-2sin(s+2t)$
$v_{ts}=-2rcos(s+2t)$
$v_{tt}=-4rcos(s+2t)$
Work Step by Step
Given: $v=rcos(s+2t)$
Partial differentiate with respect to $r$ is $v_r=cos(s+2t)$ ...(1)
Partial differentiate with respect to $s$ is $v_s=-rsin(s+2t)$ ...(2)
Partial differentiate with respect to $t$ is $v_t=-2rsin(s+2t)$ ...(3)
Partial differentiate equation (1) with respect to $r$ is
$v_{rr}=0$
Partial differentiate equation (1) with respect to $s$ is$v_{rs}=-sin(s+2t)$
Partial differentiate equation (1) with respect to $t$ is
$v_{rt}=-2sin(s+2t)$
Partial differentiate equation (2) with respect to $r$ is$v_{sr}=-sin(s+2t)$
Partial differentiate equation (2) with respect to $s$ is$v_{ss}=-rcos(s+2t)$
Partial differentiate equation (2) with respect to $t$ is$v_{st}=-2rcos(s+2t)$
Partial differentiate equation (3) with respect to $r$ is$v_{tr}=-2sin(s+2t)$
Partial differentiate equation (3) with respect to $s$ is$v_{ts}=-2rcos(s+2t)$
Partial differentiate equation (3) with respect to $t$ is$v_{tt}=-4rcos(s+2t)$