Calculus and Its Applications, Chapter 1, 1.4, Section 1.4, Problem 40

Determine the $f'(x)$ of the function $\displaystyle f(x) = \frac{1 }{\sqrt{x}}$

$
\begin{equation}
\begin{aligned}
\frac{f(x + h) - f(x)}{h} &= \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h}\\
\\
&= \frac{\sqrt{x} - \sqrt{x + h}}{h\sqrt{x}\sqrt{x + h}}
&& \text{Get the LCD}\\
\\
&= \frac{\sqrt{x} - \sqrt{x + h}}{h (\sqrt{x})( \sqrt{x + h})} \left( \frac{\sqrt{x} + \sqrt{x + h}}{\sqrt{x} + \sqrt{x + h}} \right)
&& \text{Multiply 1 by using the conjugate of the numerator}\\
\\
&= \frac{x - x - h}{h[x (\sqrt{x + h}) + (x + h) ( \sqrt{x})}\\
\\
&= \frac{-1}{x \sqrt{x + h} + (x + h)(\sqrt{x})}
\end{aligned}
\end{equation}
$

Thus,

$
\begin{equation}
\begin{aligned}
f'(x) = \lim_{h \to 0} \frac{ f(x + h) - f(x) }{h} &= \frac{-1}{x\sqrt{x + 0} + (x + 0)\sqrt{x}}\\
\\
&= \frac{-1}{x \sqrt{x} + x \sqrt{x}}\\
\\
&= \frac{-1}{2x \sqrt{x}}
\end{aligned}
\end{equation}
$