Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 84

Differentiate $s = \sqrt[4]{t^4 + 3t^2 + 8} \cdot 3t$

By using Product Rule and Chain Rule, we get

$
\begin{equation}
\begin{aligned}
s'(t) &= \left( t^4 + 3t^2 + 8 \right)^{\frac{1}{4}} \cdot \frac{d}{dt} (3t) + (3t) \cdot \frac{d}{dt} \left( t^4 + 3t^2 + 8 \right)^{\frac{1}{4}}\\
\\
s'(t) &= \left( t^4 +3t^2 + 8 \right)^{\frac{1}{4}} (3) + 3t \cdot \frac{1}{4} \left( t^4 + 3t^2 + 8 \right)^{\frac{1}{4} - 1} \cdot
\frac{d}{dt} \left( t^4 + 3t^2 + 8 \right)\\
\\
s'(t) &= 3 \left( t^4 +3t^2 + 8 \right)^{\frac{1}{4}} + \frac{3t}{4} \left( t^4 +3t^2 + 8 \right)^{-\frac{3}{4}} (4t^3 + 6t)\\
\\
s'(t) &= 3\left( t^4 +3t^2 + 8 \right)^{\frac{1}{4}} + \frac{3t^4 + \frac{9t^2}{2}}{\left( t^4 +3t^2 + 8 \right)^{\frac{3}{4}}}\\
\\
s'(t) &= \frac{3\left( t^4 +3t^2 + 8 \right) + 3t^4 + \frac{9}{2}t^2 }{\left( t^4 +3t^2 + 8 \right)^{\frac{3}{4}}}\\
\\
s'(t) &= \frac{3t^4 + 9t^2 + 24 + 3t^4 + \frac{9}{2}t^2 }{\left( t^4 +3t^2 + 8 \right)^{\frac{3}{4}}}\\
\\
s'(t) &= \frac{6t^4 + \frac{27}{2}t^2 + 24 }{\left( t^4 +3t^2 + 8 \right)^{\frac{3}{4}}}
\end{aligned}
\end{equation}
$