A(t) is a 2×2 matrix of differentiable functions and X(t) is a 2×1 column matrix...

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
${\displaystyle \frac{d}{dt}\left[A\left(t\right)X\left(t\right)\right]=A^{\prime }\left(t\right)X\left(t\right)+A\left(t\right)X^{\prime }\left(t\right)}$
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,
${\displaystyle \frac{d}{dt}\left[A\left(t\right)X\left(t\right)\right]}$
$=\left[\begin{array}{cc}\frac{d}{dt}{A}_{11}(t)x_1(t)& A_{12}(t)x_2(t)\\ \frac{d}{dt}{A}_{21}(t)x_1(t)& A_{22}(t)x_2(t)\end{array}\right]$
$=\left[\begin{array}{cc}\frac{d}{dt}[A_{11}(t)x_1(t)& A_{12}(t)x_2(t)]\\ \frac{d}{dt}[A_{21}(t)x_1(t)& A_{22}(t)x_2(t)]\end{array}\right]$
$=\left[\begin{array}{cc}{A}_{11}^{\prime }(t)x_1(t)+A_{11}(t)x_1^{\prime }(t)+& A_{12}^{\prime }(t)x_2(t)+A_{12}(t)x_2^{\prime }(t)\\ A_{21}^{\prime }(t)x_1(t)+A_{21}(t)x_1^{\prime }(t)+& A_{22}^{\prime }(t)x_2(t)+A_{22}(t)x_2^{\prime }(t)\end{array}\right]$
Step 3
Now, $A^{\prime }\left(\begin{array}{c}t\end{array}\right)=\left[\begin{array}{cc}{A}_{11}^{\prime }(t)& A_{12}^{\prime }(t)\\ A_{21}(t)& A_{22}(t)\end{array}\right]$