If $y=\frac{p x+q}{r x+s}$, then value of $2 y_1 y_3$ is: (where $y_1=y', y_2=y''$ and $\left.y_3=y'''\right)$

$3\left(y_2\right)^2$

The correct answer is Option (3) → $3\left(y_2\right)^2$

$y=\frac{px+q}{rx+s}$

$y'=\frac{p(rx+s)-r(px+q)}{(rx+s)^2}=\frac{ps-qr}{(rx+s)^2}$

$y''=\frac{-2r(ps-qr)}{(rx+s)^3}$

$y'''=\frac{6r^2(ps-qr)}{(rx+s)^4}$

$2y_1y_3=2\cdot \frac{ps-qr}{(rx+s)^2}\cdot \frac{6r^2(ps-qr)}{(rx+s)^4}$

$= \frac{12r^2(ps-qr)^2}{(rx+s)^6}$

$y_2^2=\left(\frac{-2r(ps-qr)}{(rx+s)^3}\right)^2=\frac{4r^2(ps-qr)^2}{(rx+s)^6}$

$2y_1y_3=3y_2^2$

$2y_1y_3 = 3y_2^2$