Answer
$$D'(t)=\frac{-3-16t^2}{64t^4}$$
Work Step by Step
$$D(t)=\frac{1+16t^2}{(4t)^3}$$
$$D(t)=\frac{1+16t^2}{64t^3}$$
$$D(t)=\frac{1}{64t^3}+\frac{1}{4t}$$
$$D(t)=\frac{1}{64}t^{-3}+\frac{1}{4}t^{-1}$$
Therefore, the derivative of $D(t)$ is $$D'(t)=\frac{1}{64}\frac{d}{dt}(t^{-3})+\frac{1}{4}\frac{d}{dt}(t^{-1})$$
$$D'(t)=\frac{1}{64}\times(-3)t^{-4}+\frac{1}{4}\times(-1)t^{-2}$$
$$D'(t)=\frac{-3}{64t^4}-\frac{1}{4t^2}$$
$$D'(t)=\frac{-3-16t^2}{64t^4}$$
