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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
If $y = {x^{({x^x})}}$, then ${{dy} \over {dx}} = $
  • A
    $y[{x^x}(\log ex).\log x + {x^x}]$
  • B
    $y[{x^x}(\log ex).\log x + x]$
  • ✓
    $y[{x^x}(\log ex).\log x + {x^{x - 1}}]$
  • D
    $y[{x^x}({\log _e}x).\log x + {x^{x - 1}}]$
✓

Answer

Correct option: C.
$y[{x^x}(\log ex).\log x + {x^{x - 1}}]$
c
(c) $y = {x^{({x^x})}} \Rightarrow \log y = {x^x}\log x$

==> $\frac{1}{y}\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}}.\log x + \frac{1}{x}.z$ , (where ${x^x} = z$)

$ \Rightarrow \frac{{dy}}{{dx}} = {x^{({x^x})}}\left[ {{x^x}(\log ex).\log x + {x^{x - 1}}} \right]$,      $\left\{ \because \frac{dz}{dx}={{x}^{x}}\log ex \right\}$.

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