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If $(x e)^{y}=e^{y}$, then $\frac{d y}{d x}$ is
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Verified Answer
The correct answer is:
$\frac{\log x}{(1+\log x)^{2}}$
We have, $(x e)^{y}=e^{x}$
Taking log on both sides at base $e$, we get
$\begin{aligned}
y \log (x e) &=x \log e \\
\Rightarrow \quad & y(\log x+\log e) &=x\left(\because \log _{e} e=1\right) \\
\Rightarrow \quad y &=\frac{x}{\log x+1}
\end{aligned}$
On differentiating both sides w.r.t. $x$, we get
$\begin{aligned}
\frac{d y}{d x} &=\frac{(\log x+1)-x\left(\frac{1}{x}+0\right)}{(\log x+1)^{2}} \\
\Rightarrow \quad \frac{d y}{d x} &=\frac{\log x}{(\log x+1)^{2}}
\end{aligned}$
Taking log on both sides at base $e$, we get
$\begin{aligned}
y \log (x e) &=x \log e \\
\Rightarrow \quad & y(\log x+\log e) &=x\left(\because \log _{e} e=1\right) \\
\Rightarrow \quad y &=\frac{x}{\log x+1}
\end{aligned}$
On differentiating both sides w.r.t. $x$, we get
$\begin{aligned}
\frac{d y}{d x} &=\frac{(\log x+1)-x\left(\frac{1}{x}+0\right)}{(\log x+1)^{2}} \\
\Rightarrow \quad \frac{d y}{d x} &=\frac{\log x}{(\log x+1)^{2}}
\end{aligned}$
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