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If \( x^y=e^{x-y}\), show that \( \large\frac{dy}{dx}=\frac{logx}{\{log(xe)\}^2} \)

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  • $\log m^n=n\log m$
  • $\large\frac{d}{dx}$$\log x=\large\frac{1}{x}$
Step 1:
But $x^y=e^{\Large \log x^y}=e^{\Large y\log x}$
$e^{\Large y\log x}=e^{\Large x-y}$
$\Rightarrow y\log x=x-y$
$y\log x+y=x$
$y(1+\log x)=x$
$y=\large\frac{x}{1+\log x}$
Step 2:
Diff w.r.t $x$ on both sides we get,
$\large\frac{dy}{dx}=\frac{(1+\log x).1-x(\Large\frac{1}{x})}{(1+\log x)^2}$
$\qquad=\large\frac{\log x}{(1+\log x)^2}$
$\qquad=\large\frac{\log x}{(1+\log xe)^2}$
Hence proved.
answered Nov 12, 2013 by sreemathi.v

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