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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $f(y) = \sin^{-1}\left(\frac{3y}{2} - \frac{y^3}{2}ight)$.We can rewrite the expression inside the inverse sine function as:$f(y) = \sin^{-1}\left(3\left(\frac{y}{2}ight) - 4\left(\frac{y}{2}ight)^3ight)$.Using the trigonometric identity $\sin(3	heta) = 3\sin	heta - 4\sin^3	heta$,let $\sin	heta = \frac{y}{2}$,which implies $	heta = \sin^{-1}\left(\frac{y}{2}ight)$.Then,$f(y) = \sin^{-1}(\sin(3	heta)) = 3	heta = 3\sin^{-1}\left(\frac{y}{2}ight)$.Now,differentiate with respect to $y$:$\frac{d}{dy}f(y) = 3 \cdot \frac{d}{dy}\left(\sin^{-1}\left(\frac{y}{2}ight)ight)$.Using the chain rule $\frac{d}{dx}\sin^{-1}(u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$:$\frac{d}{dy}f(y) = 3 \cdot \frac{1}{\sqrt{1 - (y/2)^2}} \cdot \frac{1}{2}$.$\frac{d}{dy}f(y) = \frac{3}{2 \sqrt{1 - y^2/4}} = \frac{3}{2 \sqrt{\frac{4 - y^2}{4}}} = \frac{3}{2 \cdot \frac{\sqrt{4 - y^2}}{2}} = \frac{3}{\sqrt{4 - y^2}}$.

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

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