Step1

Let The matrices A(t) and X(t) be

$A\left(\begin{array}{c}t\end{array}\right)=\left[\begin{array}{cc}{A}_{11}(t)& {A}_{12}(t)\\ {A}_{21}(t)& {A}_{22}(t)\end{array}\right]$

$X\left(\begin{array}{c}t\end{array}\right)=\left[\begin{array}{cc}{x}_{1}(t)& \\ {x}_{2}(t)\end{array}\right]$

The product rule of differentiation states that

$\frac{d}{dt}\left[A\left(t\right)X\left(t\right)\right]={A}^{\prime}\left(t\right)X\left(t\right)+A\left(t\right){X}^{\prime}\left(t\right)$

Step2

$A\left(\begin{array}{c}t\end{array}\right)X\left(\begin{array}{c}t\end{array}\right)=\left[\begin{array}{cc}{A}_{11}(t)& {A}_{12}(t)\\ {A}_{21}(t)& {A}_{22}(t)\end{array}\right]\left[\begin{array}{cc}{x}_{1}(t)& \\ {x}_{2}(t)\end{array}\right]$

$=\left[\begin{array}{cc}{A}_{11}(t){x}_{1}(t)& {A}_{12}(t){x}_{2}(t)\\ {A}_{21}(t){x}_{1}(t)& {A}_{22}(t){x}_{2}(t)\end{array}\right]$

Now,

$\frac{d}{dt}\left[A\left(t\right)X\left(t\right)\right]$

$=\left[\begin{array}{cc}\frac{d}{dt}{A}_{11}(t){x}_{1}(t)& {A}_{12}(t){x}_{2}(t)\\ \frac{d}{dt}{A}_{21}(t){x}_{1}(t)& {A}_{22}(t){x}_{2}(t)\end{array}\right]$

$=\left[\begin{array}{cc}\frac{d}{dt}[{A}_{11}(t){x}_{1}(t)& {A}_{12}(t){x}_{2}(t)]\\ \frac{d}{dt}[{A}_{21}(t){x}_{1}(t)& {A}_{22}(t){x}_{2}(t)]\end{array}\right]$

$=\left[\begin{array}{cc}{A}_{11}^{\prime}(t){x}_{1}(t)+{A}_{11}(t){x}_{1}^{\prime}(t)+& {A}_{12}^{\prime}(t){x}_{2}(t)+{A}_{12}(t){x}_{2}^{\prime}(t)\\ {A}_{21}^{\prime}(t){x}_{1}(t)+{A}_{21}(t){x}_{1}^{\prime}(t)+& {A}_{22}^{\prime}(t){x}_{2}(t)+{A}_{22}(t){x}_{2}^{\prime}(t)\end{array}\right]$

Step 3

Now, ${A}^{\prime}\left(\begin{array}{c}t\end{array}\right)=\left[\begin{array}{cc}{A}_{11}^{\prime}(t)& {A}_{12}^{\prime}(t)\\ {A}_{21}(t)& {A}_{22}(t)\end{array}\right]$