Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 30

Determine the $y'$ of the function $\displaystyle y = 3x^{\frac{4}{3}} - x^{\frac{1}{2}}$
By using Chain Rule,

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


Then,

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