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${a \over {\sqrt {1 - {a^2}{x^2}} }}$
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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