If $A=\begin{bmatrix}f(x) & g(x)\\h(x)& t(x)\end{bmatrix}$ Then $\frac{d}{dx}|A|$ is :

$\begin{vmatrix}f'(x) & g'(x)\\h(x) & t(x)\end{vmatrix}+\begin{vmatrix}f(x) & g(x)\\h'(x) & t'(x)\end{vmatrix}$

The correct answer is Option (4) → $\begin{vmatrix}f'(x) & g'(x)\\h(x) & t(x)\end{vmatrix}+\begin{vmatrix}f(x) & g(x)\\h'(x) & t'(x)\end{vmatrix}$

$\frac{d}{dx}|A|=\frac{d}{dx}(f(x)t(x)-g(x)h(x))$

Using product rule,

$\frac{d}{dx}f(x)t(x)=f'(x)t(x)+f(x)t'(x)$

Similarly

$\frac{d}{dx}(g(x)h(x))=g'(x)h(x)+g(x)h'(x)$

$∴\frac{d}{dx}|A|=(f'(x)t(x)+f(x)t'(x))-(g'(x)h(x)+g(x)h'(x))$

$∴\begin{vmatrix}f'(x) & g'(x)\\h(x) & t(x)\end{vmatrix}+\begin{vmatrix}f(x) & g(x)\\h'(x) & t'(x)\end{vmatrix}$