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