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If $x^{x}=y^{y}$, then $\frac{d y}{d x}$ is
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Verified Answer
The correct answer is:
$\frac{1+\log x}{1+\log y}$
We have,
$x^{x}=y^{y} \Rightarrow x \log x=y \log y$
$\Rightarrow \quad 1 \cdot \log x+\frac{x}{x}=y^{\prime} \log y+\frac{y}{y} y^{\prime}$
$\Rightarrow \quad \log x+1=y^{\prime}(\log y+1)$
$\Rightarrow \quad \frac{d y}{d x}=\frac{1+\log x}{1+\log y}$
$x^{x}=y^{y} \Rightarrow x \log x=y \log y$
$\Rightarrow \quad 1 \cdot \log x+\frac{x}{x}=y^{\prime} \log y+\frac{y}{y} y^{\prime}$
$\Rightarrow \quad \log x+1=y^{\prime}(\log y+1)$
$\Rightarrow \quad \frac{d y}{d x}=\frac{1+\log x}{1+\log y}$
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