Calculus: Early Transcendentals, Chapter 3, 3.1, Section 3.1, Problem 24

Differentiate the function $\displaystyle \frac{x^2 - 2\sqrt{x}}{x}$


$
\begin{equation}
\begin{aligned}
\text{We have } y &= \frac{x^2 - 2x^{\frac{1}{2}}}{x} \\
\\
&= \frac{x^2}{x} - \frac{2x^{\frac{1}{2}}}{x}\\
\\
&= x^{2 - 1} - 2x^{\frac{1}{2} - 1}\\
\\
&= x - 2x^{-\frac{1}{2}}
\end{aligned}
\end{equation}
$


So,

$
\begin{equation}
\begin{aligned}
\frac{dy}{dx} &= \frac{d}{dx} \left( x - 2x^{-\frac{1}{2}} \right)\\
\\
&= \frac{d}{dx} (x) - 2 \cdot \frac{d}{dx} \left( x^{-\frac{1}{2}} \right)\\
\\
&= (1) - 2 \cdot \left( - \frac{1}{2} \right) x^{-\frac{1}{2} - 1}\\
\\
&= 1 + x^{-\frac{3}{2}} \text{ or } 1 + \frac{1}{x^{\frac{3}{2}}}
\end{aligned}
\end{equation}
$