Answer
$D'(t)=\dfrac{-16t^2-3}{64t^4}$
Work Step by Step
$D(t)=\dfrac{1+16t^2}{(4t)^3}$
Use the quotient rule:
$D'(t)=\dfrac{(64t^3)(0+32t)-(1+16t^2)(3)(4t)^2(4)}{(4t)^6}$
Apply exponents:
$D'(t)=\dfrac{2048t^4-(1+16t^2)(3)(16t^2)(4)}{4096t^6}$
Perform multiplication:
$\begin{align}
D'(t)&=\dfrac{2048t^4-(1+16t^2)(192t^2)}{4096t^6}\\&=\dfrac{2048t^4-192t^2-3072t^4}{4096t^6}
\end{align}$
Combine like terms:
$D'(t)=\dfrac{-1024t^4-192t^2}{4096t^6}$
Factor:
$D'(t)=\dfrac{(64t^2)(-16t^2-3)}{4096t^6}$
Simplify:
$D'(t)=\dfrac{-16t^2-3}{64t^4}$
