If $Δ_1=\begin{vmatrix}x & b & b \\a & x & b\\a & a & x\end{vmatrix} $ and $Δ_2=\begin{vmatrix} x & b \\ a& x \end{vmatrix}$ are determinants, then :

$Δ_1=3(Δ_2)^2$ $\frac{d}{dx}(Δ_1)=3(Δ_2)^2$ $\frac{d}{dx}(Δ_1)=3Δ_2$ $Δ_1=3(Δ_2)^{\frac{3}{2}}$

$\frac{d}{dx}(Δ_1)=3Δ_2$

The correct answer is Option (3) → $\frac{d}{dx}(Δ_1)=3Δ_2$ $Δ_2=x^2-ab$ $Δ_1=x(x^2-ab)+b(ab-ax)+b(a^2-ax)$ $=x^3-abx+ab^2-abx+a^2b-abx$ $\frac{dΔ_1}{dx}=3x^2-ab-ab-ab=3(x^2-ab)$ $\frac{dΔ_1}{dx}=3Δ_2$