Solution
= Solution
Differentiating each <Boltzmann factor> yields
$$
\boxed{\frac{\partial q_3}{\partial T}
=\frac{\theta_1e^{-\theta_1/T}+\theta_2e^{-\theta_2/T}}{T^2}}.
$$
The derivative is positive: warming increases occupation of both excited <energy levels>. At $100\ \mathrm K$,
$$
\boxed{\left.\frac{\partial q_3}{\partial T}\right|_{100\,\mathrm K}
=1.56692\times10^{-3}\ \mathrm K^{-1}}.
$$