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5. continuity and differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths5. continuity and differentiation1 Mark
MCQ
${d \over {dx}}\left( {{{{{\cot }^2}x - 1} \over {{{\cot }^2}x + 1}}} \right) = $
  • A
    $ - \sin 2x$
  • B
    $2\sin 2x$
  • C
    $2\cos 2x$
  • ✓
    $ - 2\sin 2x$
✓

Answer

Correct option: D.
$ - 2\sin 2x$
d
(d) $\frac{d}{{dx}}\left[ {\frac{{{{\cot }^2}x - 1}}{{{{\cot }^2}x + 1}}} \right] = \frac{d}{{dx}}\left[ {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x + {{\sin }^2}x}}} \right]$

$ = \frac{d}{{dx}}[\cos 2x] = - 2\sin 2x$.

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