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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 $y = {x^{\sqrt x }},$then ${{dy} \over {dx}} =$
  • ✓
    ${x^{\sqrt x }}{{2 + \log x} \over {2\sqrt x }}$
  • B
    ${x^{\sqrt x }}{{2 + \log x} \over {\sqrt x }}$
  • C
    ${{2 + \log x} \over {2\sqrt x }}$
  • D
    None of these
✓

Answer

Correct option: A.
${x^{\sqrt x }}{{2 + \log x} \over {2\sqrt x }}$
a
(a) $y = {x^{\sqrt x }} \Rightarrow {\log _e}y = \sqrt x \log x$

==> $\frac{1}{y}\frac{{dy}}{{dx}} = \sqrt x \frac{1}{x} + \frac{1}{{2\sqrt x }}\log x$ or 

$\frac{{dy}}{{dx}} = {x^{\sqrt x }}\left[ {\frac{{2 + {{\log }_e}x}}{{2\sqrt x }}} \right]$

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