If Δ1=x b ba x ba a x and Δ2=x ba x then

dΔ1dx=1 b b0 x b0 a x+x0ba1ba0x+x b 0a x 0a b 1=xbax+xbax+xbax=3Δ2

If Δ1=x b ba x ba a x and Δ2=x ba x then

dΔ1dx=1 b b0 x b0 a x+x0ba1ba0x+x b 0a x 0a b 1=xbax+xbax+xbax=3Δ2

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If $Δ_{1} = |\begin{array}{lll} x b b \\ a x b \\ a a x \end{array}|$ and $Δ_{2} = |\begin{array}{ll} x b \\ a x \end{array}|$ then

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$Δ_{1} = 3 {(Δ_{2})}^{2}$

$\frac{d}{d x} (Δ_{1}) = 3 Δ_{2}$

$\frac{d}{d x} (Δ_{1}) = 3 Δ_{2}^{2}$

$Δ_{1} = 3 Δ_{2}^{3 / 2}$

answer is B.

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Detailed Solution

$\frac{d Δ_{1}}{d x} = |\begin{array}{lll} 1 b b \\ 0 x b \\ 0 a x \end{array}| + |\begin{array}{ccc} x & 0 & b \\ a & 1 & b \\ a & 0 & x \end{array}| + |\begin{array}{lll} x b 0 \\ a x 0 \\ a b 1 \end{array}|$

$\begin{aligned} = |\begin{array}{ll} x & b \\ a & x \end{array}| + |\begin{array}{ll} x & b \\ a & x \end{array}| + |\begin{array}{ll} x & b \\ a & x \end{array}| \\ = 3 Δ_{2} \end{aligned}$