Answer
$\frac{dv}{dt}=\frac{1}{6}$
Work Step by Step
given ${r+s^2+v^3=12}$,$\frac{dr}{dt}=4$ and $\frac{ds}{dt}=-3$
on differentiating the above equation with respect to t:
$\frac{dr}{dt}+2s\frac{ds}{dt}+3v^2\frac{dv}{dt}=0$
from above data:
$4+2s(-3)+3v^2\frac{dv}{dt}=0$
on putting value of $r=3$ and $s=1$
${r+s^2+v^3=12}=3+1+v^3$
so $v=2$
$4+2\times1(-3)+3\times 2^2\frac{dv}{dt}=0$
$\frac{dv}{dt}=\frac{1}{6}$
