frac{d}{{dy}}left( {{{sin }^{ - 1}}left( {fra | vedclass

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Question

(A) Let $P = \sin^{-1}\left(\frac{3y}{2} - \frac{y^3}{2}\right)$.We can rewrite the expression inside the inverse sine function as $3\left(\frac{y}{2}

Vedclass · Accepted Answer

$\frac{3}{{\sqrt {4 - {y^2}} }}$

Vedclass · Answer

(A) Let $P = \sin^{-1}\left(\frac{3y}{2} - \frac{y^3}{2}ight)$.We can rewrite the expression inside the inverse sine function as $3\left(\frac{y}{2}ight) - 4\left(\frac{y}{2}ight)^3$.Let $\frac{y}{2} = \sin 	heta$,which implies $	heta = \sin^{-1}\left(\frac{y}{2}ight)$.Substituting this into the expression,we get $P = \sin^{-1}(3\sin 	heta - 4\sin^3 	heta)$.Using the trigonometric identity $\sin(3	heta) = 3\sin 	heta - 4\sin^3 	heta$,we have $P = \sin^{-1}(\sin 3	heta) = 3	heta$.Substituting back for $	heta$,we get $P = 3\sin^{-1}\left(\frac{y}{2}ight)$.Differentiating with respect to $y$,we get $\frac{dP}{dy} = 3 \cdot \frac{1}{\sqrt{1 - (y/2)^2}} \cdot \frac{d}{dy}\left(\frac{y}{2}ight)$.$\frac{dP}{dy} = 3 \cdot \frac{1}{\sqrt{1 - y^2/4}} \cdot \frac{1}{2} = 3 \cdot \frac{1}{\sqrt{(4 - y^2)/4}} \cdot \frac{1}{2} = 3 \cdot \frac{1}{\frac{\sqrt{4 - y^2}}{2}} \cdot \frac{1}{2} = \frac{3}{\sqrt{4 - y^2}}$.

$\frac{d}{{dy}}\left( {{{\sin }^{ - 1}}\left( {\frac{{3y}}{2} - \frac{{{y^3}}}{2}} \right)} \right) = $

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