If x^y = e^ x - y , then dy dx = - Vidyadip

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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