Question

If $y=\left(\frac{ax+b}{cx+d}\right)$ , then $2\frac{dy}{dx}.\frac{d^{3}y}{d{x}^3}$ is equal to

• A. ${\left(\frac{d^{2}y}{d{x}^2}\right)}^2$
• B. $3\frac{d^{2}y}{d{x}^2}$
• C. $3{\left(\frac{d^{2}y}{d{x}^2}\right)}^2$
• D. $3\frac{d^{2}x}{d{y}^2}$

Answer 1

The correct option is C $3{\left(\frac{d^{2}y}{d{x}^2}\right)}^2$

$\because y=\left(\frac{ax+b}{cx+d}\right)orcxy+dy=ax+b$
Differentiating both sides w.r.t.x, then
$c\{x\frac{dy}{dx}+y.1\}+d\frac{dy}{dx}=aorx\frac{dy}{dx}+y+\left(\frac{d}{c}\right)\frac{dy}{dx}=\left(\frac{a}{c}\right)$
Again differentiating both sides w.r.t.x, then
$orx\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+\frac{dy}{dx}+\left(\frac{d}{c}\right)\frac{d^{2}y}{d{x}^2}=0orx+\frac{2\frac{dy}{dx}}{\left(\frac{d^{2}y}{d{x}^2}\right)}+\frac{d}{c}=0$
Again differentiating both sides w.r.t.x, then
$1+2\frac{(\frac{d^{2}y}{d{x}^2}.\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}.\frac{d^{3}y}{d{x}^3})}{{\left(\frac{d^{2}y}{d{x}^2}\right)}^2}+0=0\therefore 2\frac{dy}{dx}.\frac{d^{3}y}{d{x}^3}=3{\left(\frac{d^{2}y}{d{x}^2}\right)}^2$