Answer
$r^{\prime}(t)=\displaystyle \frac{2(5t-6)^{3}(15t^{2}+18t+40)}{(3t^{2}+4)^{2}}$
Work Step by Step
$r(t)=\displaystyle \frac{(5t-6)^{4}}{3t^{2}+4}= \frac{u(x)}{v(x)}$
Use the quotient rule
$r^{\prime}(t)=\displaystyle \frac{v(t)\cdot u^{\prime}(t)-u(t)\cdot v^{\prime}(t)}{[v(t)]^{2}}$
For $u^{\prime}(t)$, we need the chain rule:
$u^{\prime}(t)=[(5t-6)^{4}]^{\prime} =4(5t-6)^{3}\cdot(5t-6)^{\prime}=4(5t-6)^{3}\cdot 5$
$r^{\prime}(t)=\displaystyle \frac{(3t^{2}+4)[4(5t-6)^{3}\cdot 5]-(5t-6)^{4}(6t)}{(3t^{2}+4)^{2}}$
$=\displaystyle \frac{20(3t^{2}+4)(5t-6)^{3}-6t(5t-6)^{4}}{(3t^{2}+4)^{2}} \color{blue}{\text{... factor out }2(5t-6)^{3}}$
$=\displaystyle \frac{2(5t-6)^{3}[10(3t^{2}+4)-3t(5t-6)]}{(3t^{2}+4)^{2}}$
$=\displaystyle \frac{2(5t-6)^{3}(30t^{2}+40-15t^{2}+18t)}{(3t^{2}+4)^{2}}$
$r^{\prime}(t)=\displaystyle \frac{2(5t-6)^{3}(15t^{2}+18t+40)}{(3t^{2}+4)^{2}}$
