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If \( x^{y}=e^{x-y} \) then \( \frac{d y}{d x} \) is equal to
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Verified Answer
The correct answer is:
\( \frac{\log x}{(1+\log x)^{2}} \)
Given that, \( x^{y}=e^{x-y} \)
Taking log both sides, we get
\[
\begin{array}{l}
y \log x=x-y \\
y(1+\log x)=x \\
\Rightarrow y=\frac{x}{(1+\log x)}
\end{array}
\]
\[
\begin{array}{l}
\text { So, } \frac{d y}{d x}=\frac{(1+\log x)-x\left(\frac{1}{x}\right)}{(1+\log x)^{2}} \\
=\frac{\log x}{(1+\log x)^{2}}
\end{array}
\]
Taking log both sides, we get
\[
\begin{array}{l}
y \log x=x-y \\
y(1+\log x)=x \\
\Rightarrow y=\frac{x}{(1+\log x)}
\end{array}
\]
\[
\begin{array}{l}
\text { So, } \frac{d y}{d x}=\frac{(1+\log x)-x\left(\frac{1}{x}\right)}{(1+\log x)^{2}} \\
=\frac{\log x}{(1+\log x)^{2}}
\end{array}
\]
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