If y = (x x)^ , x , then dy dx = - Vidyadip

MCQ

If $y = {(x\log x)^{\log \,\log x}}$, then ${{dy} \over {dx}} = $

✓
${(x\log x)^{\log \log x}}\left\{ {{1 \over {x\log x}}(\log x + \log \log x) + (\log \,\,\log x){\rm{ }}\left( {{1 \over x} + {1 \over {x\log x}}} \right){\rm{ }}} \right\}$
B
${(x\log x)^{x\log x}}\log \log x\left[ {{2 \over {\log x}} + {1 \over x}} \right]$
C
${(x\log x)^{x\log x}}\log \log x\left[ {{2 \over {\log x}} + {1 \over x}} \right]$
D
None of these

✓

Answer

Correct option: A.

${(x\log x)^{\log \log x}}\left\{ {{1 \over {x\log x}}(\log x + \log \log x) + (\log \,\,\log x){\rm{ }}\left( {{1 \over x} + {1 \over {x\log x}}} \right){\rm{ }}} \right\}$

a
(a) $y = {(x\log x)^{\log \log x}}$

==> $\log y = \log \log x[\log x + \log \log x]$

==> $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{x\log x}}(\log x + \log \log x) + \log \log x\left( {\frac{1}{x} + \frac{1}{{x\log x}}} \right)$

==> $\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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