Answer
$p'(t)=\displaystyle \frac{2(2t+3)^{2}(4t^{2}-12t-3)}{(4t^{2}-1)^{2}}$
Work Step by Step
$p(t)=\displaystyle \frac{(2t+3)^{3}}{4t^{2}-1}= \frac{u(x)}{v(x)}$
Use the quotient rule
$p^{\prime}(t)=\displaystyle \frac{v(t)\cdot u^{\prime}(t)-u(t)\cdot v^{\prime}(t)}{[v(t)]^{2}}$
For $u^{\prime}(t)$, we need the chain rule:
$u^{\prime}(t)=[(2t+3)^{3}]^{\prime} =3(2t+3)^{2}\cdot(2t+3)^{\prime}=3(2t+3)^{2}\cdot 2$
$p^{\prime}(t)=\displaystyle \frac{(4t^{2}-1)[3(2t+3)^{2}\cdot 2]-(2t+3)^{3}(8t)}{(4t^{2}-1)^{2}}$
$=\displaystyle \frac{6(4t^{2}-1)(2t+3)^{2}-8t(2t+3)^{3}}{(4t^{2}-1)^{2}}$
... factor out $(2t+3)^{2}$ in the numerator...
$=\displaystyle \frac{(2t+3)^{2}[6(4t^{2}-1)-8t(2t+3)]}{(4t^{2}-1)^{2}}$
$=\displaystyle \frac{(2t+3)^{2}[24t^{2}-6-16t^{2}-24t]}{(4t^{2}-1)^{2}}$
$=\displaystyle \frac{(2t+3)^{2}[8t^{2}-24t-6]}{(4t^{2}-1)^{2}}$
$=\displaystyle \frac{2(2t+3)^{2}(4t^{2}-12t-3)}{(4t^{2}-1)^{2}}$
