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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:
$\frac{1+\log x}{1+\log y}$
Given, $x^{x}=y^{y}$
Taking log on both sides, we get
$$
x \log x=y \log y
$$
Differentiating w.r.t. $y$, we get
$$
\begin{aligned}
&y \cdot \frac{1}{y} \cdot \frac{d y}{d x}+\log y \frac{d y}{d x}=x \frac{1}{x}+\log x \\
&\Rightarrow \quad \frac{d y}{d x}(1+\log y)=1+\log x \\
&\Rightarrow \quad \frac{d y}{d x}=\frac{1+\log x}{1+\log y}
\end{aligned}
$$
Taking log on both sides, we get
$$
x \log x=y \log y
$$
Differentiating w.r.t. $y$, we get
$$
\begin{aligned}
&y \cdot \frac{1}{y} \cdot \frac{d y}{d x}+\log y \frac{d y}{d x}=x \frac{1}{x}+\log x \\
&\Rightarrow \quad \frac{d y}{d x}(1+\log y)=1+\log x \\
&\Rightarrow \quad \frac{d y}{d x}=\frac{1+\log x}{1+\log y}
\end{aligned}
$$
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