Answer
$y'=\dfrac{12}{(t-2)^5}$
Work Step by Step
$y=-\dfrac{3}{(t-2)^4}=-3(t-2)^{-4}.$
Using the Chain Rule:
$u=(t-2)$; $\dfrac{du}{dt}=1$
$\dfrac{dy}{du}=\dfrac{12}{u^5}$
$\dfrac{dy}{dt}=\dfrac{dy}{du}\times\dfrac{du}{dt}=\dfrac{12}{(t-2)^5}.$
