## Thomas' Calculus 13th Edition

$-0.00005$ amp/s
As we know that $V=IR$ and $\dfrac{dV}{dt}=R\dfrac{dI}{dt}+I\dfrac{d R}{d t}$ Plug the given values respectively. $(-0.01)=(600)\dfrac{dI}{dt}+(0.04)(0.5)$ $\implies \dfrac{dI}{dt}=\dfrac{(-0.01)-(0.04)(0.5)}{600}$ Hence, $\dfrac{dI}{dt}=-0.00005$ amp/s