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 $(0, \infty)$