Answer
$$g'(t)=\frac{1}{2t\sqrt t}$$
Domain of $g(t)$ and $g'(t)$ are both $(0,\infty)$
Work Step by Step
$$g(t)=\frac{1}{\sqrt t}$$
According to definition, $$g'(t)=\lim\limits_{h\to0}\frac{g(t+h)-g(t)}{h}$$
$$g'(t)=\lim\limits_{h\to0}\frac{\frac{1}{\sqrt{t+h}}-\frac{1}{\sqrt t}}{h}$$
$$g'(t)=\lim\limits_{h\to0}\frac{\sqrt t-\sqrt{t+h}}{h\sqrt t\sqrt{t+h}}$$
Multiply both numerator and denominator by $(\sqrt t+\sqrt{t+h})$, the numerator would become $$(\sqrt t-\sqrt{t+h})(\sqrt t+\sqrt{t+h})$$
$$=t-(t+h)$$
$$=-h$$
Therefore, $$g'(t)=\lim\limits_{h\to0}\frac{-h}{h\sqrt t\sqrt{t+h}(\sqrt t+\sqrt{t+h})}$$
$$g'(t)=\lim\limits_{h\to0}\frac{-1}{\sqrt t\sqrt{t+h}(\sqrt t+\sqrt{t+h})}$$
$$g'(t)=\frac{-1}{\sqrt t\sqrt t(\sqrt t+\sqrt t)}$$
$$g'(t)=\frac{-1}{2t\sqrt t}$$
*For $g(t)$:
Since $t$ must be $\ge0$ and $\sqrt t\ne0$, which means $t\ne0$ the domain of $g(t)$ is $(0,\infty)$
*For $g'(t)$:
Since $t$ must be $\ge0$ and $\sqrt t\ne0$, which means $t\ne0$ the domain of $g(t)$ is $(0,\infty)$
