Answer
$g'(-1) = 2 \\ g'(2) = -\dfrac{1}{4} \\ g'(\sqrt 3) = -\dfrac{2}{3\sqrt 3}$
Work Step by Step
Consider $g(t) = \dfrac{1}{t^{2}}$
We will have to use definition of derivative, such as:
$g'(t) = \lim\limits_{h \to 0}\dfrac{g(t+h) - g(t)}{h}= \lim\limits_{h \to 0}\dfrac{\dfrac{1}{(t+h)^{2} - }\dfrac{1}{t^{2}}}{h}=-2(\dfrac{1}{t^{3}})$
Now,we have
$g'(-1) = 2 \\ g'(2) = -2(\dfrac{1}{(2)^{3}})=-\dfrac{1}{4} \\ g'(\sqrt 3)=-2(\dfrac{1}{(\sqrt 3)^{3}}) = -\dfrac{2}{3\sqrt 3}$
