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If $x^{y}=e^{x-y}$ then $d y / d x$ is equal to which one of the following?
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Verified Answer
The correct answer is:
$\frac{(\log x)}{(1+\log x)^{2}}$
$x^{y}=e^{x-y}$
Taking log both sides, we get $\Rightarrow y \cdot \log x=x-y$
$\Rightarrow y=\frac{x}{1+\log x}$
$\Rightarrow \frac{d y}{d x}=\frac{(1+\log x)-x\left(0+\frac{1}{x}\right)}{(1+\log x)^{2}}$
$=\frac{(1+\log x)-1}{(1+\log x)^{2}}=\frac{\log x}{(1+\log x)^{2}}$
Taking log both sides, we get $\Rightarrow y \cdot \log x=x-y$
$\Rightarrow y=\frac{x}{1+\log x}$
$\Rightarrow \frac{d y}{d x}=\frac{(1+\log x)-x\left(0+\frac{1}{x}\right)}{(1+\log x)^{2}}$
$=\frac{(1+\log x)-1}{(1+\log x)^{2}}=\frac{\log x}{(1+\log x)^{2}}$
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