Question

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

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

Answer 1

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

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