MCQ
If $y = {({x^x})^x}$, then ${{dy} \over {dx}} =$
  • A
    ${({x^x})^x}(1 + 2\log x)$
  • B
    ${({x^x})^x}(1 + \log x)$
  • $x{({x^x})^x}(1 + 2\log x)$
  • D
    $x\,{({x^x})^x}(1 + \log x)$

Answer

Correct option: C.
$x{({x^x})^x}(1 + 2\log x)$
c
(c) $y = {({x^x})^x} \Rightarrow {\log _e}y = x{\log _e}{(x)^x} = {x^2}.{\log _e}x$

==> $\frac{1}{y}\frac{{dy}}{{dx}} =  {x^2}.\frac{1}{x} + 2x.{\log _e}x$

$\therefore \frac{{dy}}{{dx}} = x{({x^x})^x}[1 + 2{\log _e}x]$.

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