d dy ( ^ - 1 ( 3y 2 - y^3 2 ) ) = - Vidyadip

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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