Question
यदि $y = {(x\log x)^{\log \,\log x}}$, तो $\frac{{dy}}{{dx}} = $
==> $\log y = \log \log x[\log x + \log \log x]$
==> $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{x\log x}}(\log x + \log \log x) + \log \log x\left( {\frac{1}{x} + \frac{1}{{x\log x}}} \right)$
==> $\frac{{dy}}{{dx}} = y\{ \frac{1}{{x\log x}}(\log x + \log \log x) + \log \log x\left( {\frac{1}{x} + \frac{1}{{x\log x}}} \right)\} $
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