Question
यदि ${x^y} = {e^{x - y}}$, तब $\frac{{dy}}{{dx}} = $
==> $y = \frac{x}{{1 + \log x}}$
==> $\frac{{dy}}{{dx}} = \log x{(1 + \log x)^{ - 2}} = \log x{[\log ex]^{ - 2}}$.
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