Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 26

Find the derivative of $\displaystyle g(t) = \frac{1}{\sqrt{t}}$ using the definition and the domain of its derivative.

Using the definition of derivative


$
\begin{equation}
\begin{aligned}

\qquad g'(t) &= \lim_{h \to 0} \frac{g(t + h) - g(t)}{h}
&&
\\
\\
\qquad g'(t) &= \lim_{h \to 0} \frac{\displaystyle \frac{1}{\sqrt{t + h}}- \frac{1}{\sqrt{t}}}{h}
&& \text{Substitute $g(t + h)$ and $g(t)$}
\\
\\
\qquad g'(t) &= \lim_{h \to 0} \frac{\sqrt{t} - \sqrt{t + h}}{(h)(\sqrt{t}) (\sqrt{t + h})}
&& \text{Get the LCD of the numerator and simplify}
\\
\\
\qquad g'(t) &= \lim_{h \to 0} \frac{\sqrt{t} - \sqrt{t + h}}{(h)(\sqrt{t})(\sqrt{t + h}) } \cdot \frac{\sqrt{t} + \sqrt{t + h}}{\sqrt{t} + \sqrt{t + h}}
&& \text{Multiply both numerator and denominator by $(\sqrt{t} + \sqrt{t + h})$}
\\
\\
\qquad g'(t) &= \lim_{h \to 0} \frac{t - (t + h)}{(h)(\sqrt{t})(\sqrt{t + h})(\sqrt{t} + \sqrt{t + h})}
&& \text{Combine like terms}
\\
\\
\qquad g'(t) &= \lim_{h \to 0} \frac{-\cancel{h}}{\cancel{(h)}(\sqrt{t})(\sqrt{t + h})(\sqrt{t} + \sqrt{t + h}) }
&& \text{Cancel out like terms}
\\
\\
\qquad g'(t) &= \lim_{h \to 0} \frac{-1}{(\sqrt{t})(\sqrt{t + h})(\sqrt{t} + \sqrt{t + h})} = \frac{-1}{(\sqrt{t})(\sqrt{t + 0})(\sqrt{t} + \sqrt{t + 0})}
&& \text{Evaluate the limit}
\\
\\
\qquad g'(t) &= \frac{-1}{(\sqrt{t})(\sqrt{t})(\sqrt{t} + \sqrt{t})}
&& \text{Simplify the equation}
\\
\\
\qquad g'(t) &= \frac{-1}{2t \sqrt{t}}

\end{aligned}
\end{equation}
$



Both functions involve square root that is continuous on $x \geq 0$. However, $\sqrt{x}$ is placed in the denominator that's why is not included. Therefore, the domain of $g(t)$ and $g'(t)$ is $(0, \infty)$