Question

If $x^y=y^x$, then $x(x-y\log x)\frac{dy}{dx}$ is equal to :

• A. $y(y-x\log y)$
• B. $y(y+x\log y)$
• C. $x(x+y\log x)$
• D. $x(y-x\log y)$

Answer 1

Answer: (A)

$\because x^y=y^x$
Taking log on both sides, we get
$y\log x=x\log y$
On differentiating with respect to $x$,
we get
$y\cdot \frac{1}{x}+\log x\frac{dy}{dx}$
$=x\cdot \frac{1}{y}\frac{dy}{dx}+\log y$
$\Rightarrow \frac{(x-y\log x)}{y}\frac{dy}{dx}$
$=\frac{-x\log y+y}{x}$
$\Rightarrow x(x-y\log x)\frac{dy}{dx}$
$=y(-x\log y+y)$