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If $x^y=y^x$, then $x(x-y \log x) \frac{d y}{d x}$ is equal to :
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Verified Answer
The correct answer is:
$y(y-x \log y)$
$\because x^y=y^x$
Taking log on both sides, we get $y \log x=x \log y$
On differentiating with respect to $x$, we get
$y \cdot \frac{1}{x}+\log x \frac{d y}{d x}=x \cdot \frac{1}{y} \frac{d y}{d x}+\log y$
$\Rightarrow \quad \frac{(x-y \log x)}{y} \frac{d y}{d x}=\frac{-x \log y+y}{x}$
$\Rightarrow \quad x(x-y \log x) \frac{d y}{d x}=y(-x \log y+y)$
Taking log on both sides, we get $y \log x=x \log y$
On differentiating with respect to $x$, we get
$y \cdot \frac{1}{x}+\log x \frac{d y}{d x}=x \cdot \frac{1}{y} \frac{d y}{d x}+\log y$
$\Rightarrow \quad \frac{(x-y \log x)}{y} \frac{d y}{d x}=\frac{-x \log y+y}{x}$
$\Rightarrow \quad x(x-y \log x) \frac{d y}{d x}=y(-x \log y+y)$
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