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