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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}}{\sin ^{ - 1}}(2ax\sqrt {1 - {a^2}{x^2}} ) = $
  • A
    ${{2a} \over {\sqrt {{a^2} - {x^2}} }}$
  • B
    ${a \over {\sqrt {{a^2} - {x^2}} }}$
  • ✓
    ${{2a} \over {\sqrt {1 - {a^2}{x^2}} }}$
  • D
    ${a \over {\sqrt {1 - {a^2}{x^2}} }}$
✓

Answer

Correct option: C.
${{2a} \over {\sqrt {1 - {a^2}{x^2}} }}$
c
(c) $\frac{d}{{dx}}{\sin ^{ - 1}}(2ax\sqrt {1 - {a^2}{x^2}} )$

Putting $ax = \sin \theta ,$ we get

$ = \frac{d}{{dx}}{\sin ^{ - 1}}[2\sin \theta \sqrt {1 - {{\sin }^2}\theta } ] = \frac{d}{{dx}}{\sin ^{ - 1}}\sin 2\theta = \frac{{2a}}{{\sqrt {1 - {a^2}{x^2}} }}$

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