Question

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

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

Answer 1

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

From the given relation $y=x\log \left(\frac{x}{a+bx}\right)$
$\frac{y}{x}=\log x-\log (a+bx)$
On differentiating w.r.t $x$, we get
$\frac{(x\frac{dy}{dx}-y)}{x^2}=\frac{1}{x}-\frac{1}{a+bx}b$
Therefore, $x\frac{dy}{dx}-y=\frac{ax}{a+bx}$ ...(i)
Again differentiating both sides w.r.t $x$ we get
$x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}-\frac{dy}{dx}=\frac{(a+bx)a-axb}{{(a+bx)}^2}$
$\Rightarrow x^3\frac{d^{2}y}{d{x}^2}=\frac{a^2{x}^2}{{(a+bx)}^2}={(x\frac{dy}{dx}-y)}^2$ ..... (from Eq(i))