Question

If $y=ln{\left(\frac{x}{a+bx}\right)}^x$,then $x^3\frac{d^{2}y}{d{x}^2}$ is equal to

• A. ${(\frac{dy}{dx}+x)}^2$
• B. ${(\frac{dy}{dx}-y)}^2$
• C. ${(x\frac{dy}{dx}+y)}^2$
• D. ${(x\frac{dy}{dx}-y)}^2$

Answer 1

The correct option is D ${(x\frac{dy}{dx}-y)}^2$

$\because y=ln{\left(\frac{x}{a+bx}\right)}^x=x(lnx-ln(a+bx\left)\right)$
$or\left(\frac{y}{x}\right)=lnx-ln(a+bx)$
Differentiating both sides w.r.t.x,then
$\frac{x\frac{dy}{dx}-y.1}{x^2}=\frac{1}{x}-\frac{b}{a+bx}=\frac{a}{x(a+bx)}\cdots \left(i\right)$
$or(x\frac{dy}{dx}-y)=\left(\frac{ax}{a+bx}\right)$
Again taking logarithm on both sides, then
$ln(x\frac{dy}{dx}-y)=ln\left(ax\right)-ln(a+bx)$
Differetiating both sides w.r.t.x, then
$\frac{x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}-\frac{dy}{dx}}{(x\frac{dy}{dx}-y)}=\frac{1}{x}-\frac{b}{a+bx}=\frac{a}{x(a+bx)}=\frac{(x\frac{dy}{dx}-y)}{x^2}[FromEq.(i\left)\right]or{x}^3\frac{d^{2}y}{d{x}^2}={(x\frac{dy}{dx}-y)}^2$