Question

${△}_1=\left|\begin{array}{ccc}x& b& b\\ a& x& b\\ a& a& x\end{array}\right|$ and $△=\left|\begin{array}{cc}x& b\\ a& x\end{array}\right|$ are the given determinants, then

• A. ${△}_1=3({△}_2{)}^2$
• B. $\frac{d}{dx}\left({△}_1\right)=3{△}_2$
• C. $\frac{d}{dx}\left({△}_1\right)=3({△}_2{)}^2$
• D. ${△}_1=3{△}_2^{3/2}$

Answer 1

The correct option is B $\frac{d}{dx}\left({△}_1\right)=3{△}_2$

${△}_1=x(x^2-ab)-b(ax-ab)+b(a^2-ax)$

$=x^3-3abx+a{b}^2+a^{2b}$

$\frac{d}{dx}\left({△}_1\right)=3x^2-3ab=3(x^2-ab)=3{△}_2$.