Does it make sense to differentiate the Arrhenius equation with respect to temperature?
Note that $A$ is a kind of a collision frequency factor, which is temperature dependent via the mean molecule speed, even if far less then the Boltzmann exponential factor.
$$\mathrm{d}k/\mathrm{d}T = \frac {\mathrm{d}A}{\mathrm{d}T} \cdot \exp{(\frac {-E_\mathrm{a}}{RT})} + A \cdot \exp{(\frac{-E_\mathrm{a}}{RT})} \cdot \frac {E_\mathrm{a}}{RT^2} \tag{1}$$
$$A = A_0 \cdot {({\frac {T}{T_0})}}^{1/2} \tag{2}$$
$$\frac {\mathrm{d}A}{\mathrm{d}T} = A_0 \cdot \frac 12 \cdot {({\frac {T}{T_0})}}^{-1/2} \cdot \frac {1}{T_0}=\frac {A_0}2 \cdot {({\frac {T_0}{T})}}^{1/2} \cdot \frac {1}{T_0} =\frac {A_0}2 \cdot \frac 1{\sqrt{T \cdot T_0}}  \tag{3}$$
By subtitution in (1), using  (2) and (3):
$$\mathrm{d}k/\mathrm{d}T = \left( \frac 12 \cdot \frac 1{\sqrt{T \cdot T_0}}   +  {({\frac {T}{T_0})}}^{1/2}  \cdot \frac {E_\mathrm{a}}{RT^2} \right) \cdot A_0 \cdot \exp{(\frac{-E_\mathrm{a}}{RT})}\tag{4}$$
As AJKOER properly noted, temperature dependance  $A=A(T)$ can be usually neglected and we can use the simplified (1), as $\frac {\mathrm{d}A}{\mathrm{d}T} = 0$:
$$\mathrm{d}k/\mathrm{d}T = A \cdot \exp{(\frac{-E_\mathrm{a}}{RT})} \cdot \frac {E_\mathrm{a}}{RT^2} \tag{1a}$$
unless the activation energy is extremely low and reaction rate is driven by diffusion ( typical one is $\ce{H3O+(aq) + OH-(aq) -&gt; 2 H2O(l)}$  ).
In solvents, we cannot usually ignore intermolecular interactions, so temperature dependence of $A$ may get quite complicated.