MCQ
If ${x^y} = {e^{x - y}}$, then ${{dy} \over {dx}} = $
  • $\log x.{[\log (ex)]^{ - 2}}$
  • B
    $\log x.{[\log (ex)]^2}$
  • C
    $\log x.{(\log x)^2}$
  • D
    None of these

Answer

Correct option: A.
$\log x.{[\log (ex)]^{ - 2}}$
a
(a) ${x^y} = {e^{x - y}}$ ==> $y\log x = x - y$

==> $y = \frac{x}{{1 + \log x}}$ 

==> $\frac{{dy}}{{dx}} = \log x{(1 + \log x)^{ - 2}} = \log x{[\log ex]^{ - 2}}$.

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