Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 26

Determine the $y'$ of the function $\displaystyle y = x^{\frac{3}{2}} - 5x$

$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} \left( x^{\frac{3}{2}} - 5x \right)\\
\\
&= \frac{d}{dx} \left( x^{\frac{3}{2}} \right) - 5 \cdot \frac{d}{dx} (x)\\
\\
&= \frac{3}{2} x^{\frac{3}{2} - 1} - 5(1)\\
\\
&= \frac{3}{2} x^{\frac{1}{2} } - 5 \text{ or } \frac{3}{2} \sqrt{x} - 5
\end{aligned}
\end{equation}
$


Then,

$
\begin{equation}
\begin{aligned}
y'' &= \frac{d}{dx} \left( \frac{3}{2} x^{\frac{1}{2}} - 5 \right)\\
\\
&= \frac{3}{2} \cdot \frac{d}{dx} \left( x^{\frac{1}{2}} \right) - \frac{d}{dx} (5)\\
\\
&= \frac{3}{2} \left( \frac{1}{2} x^{\frac{1}{2} - 1} \right) - 0 \\
\\
&= \frac{3}{4} x^{-\frac{1}{2}} \text{ or } \frac{3}{4x^{\frac{1}{2}}} \text{ or } \frac{3}{4\sqrt{x}}
\end{aligned}
\end{equation}
$