Answer
$q^{\prime}(y)=2y(y^{2}+1)^{1/4}(9y^{2}+4)$
Work Step by Step
$q(y)=4y^{2}(y^{2}+1)^{5/4}=u(y)\cdot v(y)$
Use the product rule
$\color{blue}{[u(y)\cdot v(y)]^{\prime}=u(y)\cdot v^{\prime}(y)+v(y)\cdot u^{\prime}(y)}$
For $v^{\prime}(y)$, we need the chain rule.
$\left[\begin{array}{llll}
u(y)=4y^{2} & , & v(y)=(y^{2}+1)^{5/4} & \\
u^{\prime}(y)=8y & & v^{\prime}(y)=[w(y)]^{5/4} & \\
& & v^{\prime}(y)=\frac{5}{4}\cdot[w(y)]^{5/4-1}\cdot w^{\prime}(y) & (\text{chain rule)})\\
& & v^{\prime}(y)=\frac{5}{4}(y^{2}+1)^{1/4}\cdot 2y &
\end{array}\right]$
$q^{\prime}(y)=4y^{2}\displaystyle \cdot\frac{5}{4}(y^{2}+1)^{1/4}(2y)+8y(y^{2}+1)^{5/4}$
$=10y^{3}(y^{2}+1)^{1/4}+8y(y^{2}+1)^{5/4}$
$=2y(y^{2}+1)^{1/4}[5y^{2}+4(y^{2}+1)^{4/4}]$
$=2y(y^{2}+1)^{1/4}(9y^{2}+4)$
