# Write the correct alternative in the following:

Question:

Write the correct alternative in the following:

If $\mathrm{y}=\log _{\mathrm{e}}\left(\frac{\mathrm{x}}{\mathrm{a}+\mathrm{bx}}\right)^{2}$, then $\mathrm{x}^{3} \mathrm{y}_{2}=$

A. $\left(x y_{1}-y\right)^{2}$

B. $(x+y)^{2}$

c. $\left(\frac{\mathrm{y}-\mathrm{xy}_{1}}{\mathrm{y}_{1}}\right)^{2}$

D. none of these

Solution:

Given:

$y=\left(\log _{e}\left(\frac{x}{a+b x}\right)\right)^{2}$

$=2 \log _{e}\left(\frac{x}{a+b x}\right)$

$\frac{d y}{d x}=2\left(\frac{1}{\frac{x}{a+b x}}\right)\left[\frac{a+b x-b x}{(a+b x)^{2}}\right]$

$=2\left(\frac{a+b x}{x}\right)\left[\frac{a}{(a+b x)^{2}}\right]$

$=\frac{2 a}{x(a+b x)}$

$=\frac{2 a}{\left(a x+b x^{2}\right)}$\

$x \frac{d y}{d x}=\frac{2 a x}{\left(a x+b x^{2}\right)}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2 \mathrm{a}\left\{\frac{-(\mathrm{a}+2 \mathrm{bx})}{\left(\mathrm{ax}+\mathrm{bx}^{2}\right)^{2}}\right\}$

$=(-a-2 b x) \frac{d y}{d x}$

$\mathrm{x}^{3} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\mathrm{x}^{3}(\mathrm{a}+2 \mathrm{bx}) \frac{\mathrm{dy}}{\mathrm{dx}}$