Question

If $x=y\ell n\left(xy\right)$, then $\frac{dx}{dy}$ equals-

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

Answer 1

The correct option is B $\frac{x(x+y)}{y(x-y)}$

Given that

$x=yln\left(xy\right)--------\left(1\right)$

on differentiation,

$\frac{dx}{dy}=y.\frac{d}{dy}\left(lnxy\right)+ln\left(xy\right)\frac{dy}{dy}$

(product rule)

$\frac{dx}{dy}=y.\frac{1}{xy}\frac{d}{dy}\left(xy\right)+ln\left(xy\right)$

$\frac{dx}{dy}=\frac{1}{x}.[x\frac{dy}{dy}+y\frac{dx}{dy}]+ln\left(xy\right)$

$\frac{dx}{dy}=\frac{x}{x}+\frac{y}{x}\frac{dx}{dy}+ln\left(xy\right)$

$\frac{dx}{dy}-\frac{y}{x}\frac{dx}{dy}=1+\frac{x}{y}\left[using\right(1\left)\right]$

$\frac{dx}{dy}(1-\frac{y}{x})=\frac{x+y}{y}$

$\frac{dx}{dy}\left(\frac{x-y}{x}\right)=\left(\frac{x+y}{y}\right)$

$\frac{dx}{dy}=\frac{x(x+y)}{y(x-y)}$