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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}}\left( {{1 \over {{x^4}\sec x}}} \right) = $
  • A
    ${{x\sin x + 4\cos x} \over {{x^5}}}$
  • ✓
    ${{ - (x\sin x + 4\cos x)} \over {{x^5}}}$
  • C
    ${{4\cos x - x\sin x} \over {{x^5}}}$
  • D
    None of these
✓

Answer

Correct option: B.
${{ - (x\sin x + 4\cos x)} \over {{x^5}}}$
b
(b) $\frac{d}{{dx}}\left( {\frac{1}{{{x^4}\sec x}}} \right) = \frac{d}{{dx}}\left( {\frac{{\cos x}}{{{x^4}}}} \right)$

$ = \frac{{{x^4}( - \sin x) - \cos x(4{x^3})}}{{{{({x^4})}^2}}}$

$ = \frac{{ - {x^3}(x\sin x + 4\cos x)}}{{{x^8}}} = \frac{{ - (x\sin x + 4\cos x)}}{{{x^5}}}$.

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