Answer
(a) $1.226$
(b) $5.08$
Work Step by Step
Given $$L(t)=32\left(1-\left(1+0.37 t+0.068 t^{2}+0.0085 t^{3}+0.0009 t^{4}\right)^{-1}\right)$$
(a) Since
\begin{align*}
L'(t) &= 32\left(0-\left(-\frac{0.0036t^3+0.0255t^2+0.136t+0.37}{\left(1+0.37t+0.068t^2+0.0085t^3+0.0009t^4\right)^2}\right)\right)\\
&= \frac{32\left(0.0036t^3+0.0255t^2+0.136t+0.37\right)}{\left(0.0009t^4+0.0085t^3+0.068t^2+0.37t+1\right)^2}
\end{align*}
Then
\begin{align*}
L'(6)&=\frac{32\left(0.0036(6)^3+0.0255(6)^2+0.136(6)+0.37\right)}{\left(0.0009(6)^4+0.0085(6)^3+0.068(6)^2+0.37(6)+1\right)^2}\\
&\approx1.226
\end{align*}
(b) From the following figure, we can get $$ L'(2.3) \approx 5.08$$
