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

Answer

Correct option: A.
$\frac{3}{{\sqrt {4 - {y^2}} }}$
a
$ P =\sin^{ -1}\left(3\left(\frac{y}{2}\right)-4\left(\frac{4}{2}\right)^{3}\right) $

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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