Answer
$g'(-1) = 2$
$g'(2) = -\frac{1}{4}$
$g'(\sqrt 3) = -\frac{2}{3\sqrt 3}$
Work Step by Step
$g(t) = \frac{1}{t^{2}}$
Using definition of derivative:
$g'(t) = \lim\limits_{h \to 0}\frac{g(t+h) - g(t)}{h}$
$g'(t) = \lim\limits_{h \to 0}\frac{\frac{1}{(t+h)^{2} - }\frac{1}{t^{2}}}{h}$
$g'(t) = \lim\limits_{h \to 0}\frac{-(2t+h)}{t^{4}+2t^{3}h+h^{2}t^{2}}$
$g'(t) = -2\frac{1}{t^{3}}$
Now,
$g'(-1) = 2$
$g'(2) = -\frac{1}{4}$
$g'(\sqrt 3) = -\frac{2}{3\sqrt 3}$
