## Calculus 10th Edition

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