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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
${d \over {dx}}\sqrt {{{\sec }^2}x + {\rm{cose}}{{\rm{c}}^2}x} = $
  • A
    $4\cos {\rm{ec 2}}x.\cot 2x$
  • ✓
    $ - 4\cos {\rm{ec 2}}x.\cot 2x$
  • C
    $ - 4\cos {\rm{ec }}x.\cot 2x$
  • D
    None of these
✓

Answer

Correct option: B.
$ - 4\cos {\rm{ec 2}}x.\cot 2x$
b
(b) $\frac{d}{{dx}}[\sqrt {{{\sec }^2}x + {\rm{cose}}{{\rm{c}}^2}x} ] = \frac{d}{{dx}}\left[ {\sqrt {\left( {\frac{1}{{{{\cos }^2}x}} + \frac{1}{{{{\sin }^2}x}}} \right)} } \right]$

$ = \frac{d}{{dx}}[2\,{\rm{cosec}}2x] = - 4\,{\rm{cosec}}2x\cot 2x$.

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