MCQ
$\frac{d}{{dy}}\left( {{{\sin }^{ - 1}}\left( {\frac{{3y}}{2} - \frac{{{y^3}}}{2}} \right)} \right) = $
- A$\frac{3}{{\sqrt {4 - {y^2}} }}$
- B$\frac{-3}{{\sqrt {4 - {y^2}} }}$
- C$\frac{1}{{\sqrt {4 - {y^2}} }}$
- D$\frac{-1}{{\sqrt {4 - {y^2}} }}$
Put $ \quad \frac{y}{2} =\sin \theta $
$ P =\sin ^{-1}(\sin 3 \theta)=3 \theta$
$ P =3 \sin ^{-1} \frac{y}{2} $
$ \frac{d P}{d y} =\frac{3}{\sqrt{4-y^{2}}} $
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