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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}}{\sin ^{ - 1}}(3x - 4{x^3}) = $
  • ✓
    ${3 \over {\sqrt {1 - {x^2}} }}$
  • B
    ${{ - 3} \over {\sqrt {1 - {x^2}} }}$
  • C
    ${1 \over {\sqrt {1 - {x^2}} }}$
  • D
    ${{ - 1} \over {\sqrt {1 - {x^2}} }}$
✓

Answer

Correct option: A.
${3 \over {\sqrt {1 - {x^2}} }}$
a
(a) Put $x = \sin \theta ,$ we get $\frac{d}{{dx}}{\sin ^{ - 1}}(3x - 4{x^3})$

$ = \frac{d}{{dx}}{\sin ^{ - 1}}(\sin 3\theta ) = \frac{3}{{\sqrt {1 - {x^2}} }}$.

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