## Calculus: Early Transcendentals 8th Edition

$$g'(t)=\frac{1}{2t\sqrt t}$$ Domain of $g(t)$ and $g'(t)$ are both $(0,\infty)$
$$g(t)=\frac{1}{\sqrt t}$$ 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 $(0,\infty)$