Answer
$$
W(t)=\frac{-0.02t^{2}+t}{t+1}
$$
Now, we can calculate $\frac{dy}{dt}$ by using the quotient rule. as follows:
$$
\begin{aligned} \frac{d W}{d t} &=\frac{(-0.04 t+1)(t+1)-(1)\left(-0.02 t^{2}+t\right)}{(t+1)^{2}} \\ \text { when } t=5 & \\ \frac{d W}{d t} &=\frac{(-0.2+1)(6)-(-0.5+5)}{6^{2}} \\ &=\frac{4.8-4.5}{36} \\ &=0.008 \end{aligned}
$$
Work Step by Step
$$
W(t)=\frac{-0.02t^{2}+t}{t+1}
$$
Now, we can calculate $\frac{dy}{dt}$ by using the quotient rule. as follows:
$$
\begin{aligned} \frac{d W}{d t} &=\frac{(-0.04 t+1)(t+1)-(1)\left(-0.02 t^{2}+t\right)}{(t+1)^{2}} \\ \text { If } t=5 & \\ \frac{d W}{d t} &=\frac{(-0.2+1)(6)-(-0.5+5)}{6^{2}} \\ &=\frac{4.8-4.5}{36} \\ &=0.008 \end{aligned}
$$
