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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
${d \over {dx}}{\left( {\sqrt x + {1 \over {\sqrt x }}} \right)^2} = $
  • ✓
    $1 - {1 \over {{x^2}}}$
  • B
    $1 + {1 \over {{x^2}}}$
  • C
    $1 - {1 \over {2x}}$
  • D
    None of these
✓

Answer

Correct option: A.
$1 - {1 \over {{x^2}}}$
a
(a) $\frac{d}{{dx}}{\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)^2} = \frac{d}{{dx}}\left[ {x + \frac{1}{x} + 1} \right] = 1 - \frac{1}{{{x^2}}}$.

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