Question

If $x^{y}=y^{x}$, then $x(x-y \log x) \frac{d y}{d x}$ 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{d y}{d x} $
$=x \cdot \frac{1}{y} \frac{d y}{d x}+\log y$
$\Rightarrow \frac{(x-y \log x)}{y} \frac{d y}{d x}$
$=\frac{-x \log y+y}{x}$
$\Rightarrow x(x-y \log x) \frac{d y}{d x} $
$=y(-x \log y+y)$