Answer
$$-0.1312 ~W/\Omega,\ \ \ -0.06086~W/\Omega$$
Work Step by Step
We are given
$$P=\frac{2.25 R }{(R+0.5)^{2}}$$
We find the derivative as follows:
\begin{aligned}
\frac{d P}{d R}
&=\frac{(R+0.5)^{2} \frac{d}{d R}(2.25 R)-(2.25 R) \frac{d}{d R}(R+0.5)^{2}}{(R+0.5)^{4}} \\
&=\frac{(R+0.5)^{2}(2.25)-2(2.25 R) (R+0.5)}{(R+0.5)^{4}} \\
&=\frac{(R+0.5)(2.25 R+1.125-4.5 R)}{(R+0.5)^{3}}\\
&=\frac{1.125-2.25 R}{(R+0.5)^{3}}
\end{aligned}
Now, we plug in the given values:
\begin{align*}
\left.\frac{d P}{d R}\right|_{R=3}&=\frac{1.125-2.25(3)}{((3)+0.5)^{3}}=-0.1312~W/\Omega \\
\frac{d P}{d R}\bigg |_{R=5}&=\frac{1.125-2.25(5)}{((5)+0.5)^{3}}=-0.06086~W/\Omega
\end{align*}
