If y = operatorname{Sin}^{-1}left(frac{4+5 si | vedclass

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(C) Given $y = \operatorname{Sin}^{-1}\left(\frac{4+5 \sin x}{5+4 \sin x}\right)$.Let $f(x) = \frac{4+5 \sin x}{5+4 \sin x}$.To find $\frac{dy}{dx}$,w

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$\frac{\pm 3}{5+4 \sin x}$

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(C) Given $y = \operatorname{Sin}^{-1}\left(\frac{4+5 \sin x}{5+4 \sin x}\right)$.Let $f(x) = \frac{4+5 \sin x}{5+4 \sin x}$.To find $\frac{dy}{dx}$,we use the chain rule: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - [f(x)]^2}} \cdot f'(x)$.First,calculate $f'(x)$ using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$:$u = 4+5 \sin x \implies u' = 5 \cos x$$v = 5+4 \sin x \implies v' = 4 \cos x$$f'(x) = \frac{(5 \cos x)(5+4 \sin x) - (4+5 \sin x)(4 \cos x)}{(5+4 \sin x)^2} = \frac{25 \cos x + 20 \sin x \cos x - 16 \cos x - 20 \sin x \cos x}{(5+4 \sin x)^2} = \frac{9 \cos x}{(5+4 \sin x)^2}$.Next,calculate $1 - [f(x)]^2 = 1 - \left(\frac{4+5 \sin x}{5+4 \sin x}\right)^2 = \frac{(5+4 \sin x)^2 - (4+5 \sin x)^2}{(5+4 \sin x)^2} = \frac{(25 + 16 \sin^2 x + 40 \sin x) - (16 + 25 \sin^2 x + 40 \sin x)}{(5+4 \sin x)^2} = \frac{9 - 9 \sin^2 x}{(5+4 \sin x)^2} = \frac{9 \cos^2 x}{(5+4 \sin x)^2}$.Thus,$\sqrt{1 - [f(x)]^2} = \frac{3 |\cos x|}{5+4 \sin x}$.Substituting back: $\frac{dy}{dx} = \frac{5+4 \sin x}{3 |\cos x|} \cdot \frac{9 \cos x}{(5+4 \sin x)^2} = \frac{3 \cos x}{|\cos x| (5+4 \sin x)}$.This simplifies to $\frac{3}{5+4 \sin x}$ when $\cos x > 0$ and $\frac{-3}{5+4 \sin x}$ when $\cos x < 0$. Given the options,the general derivative is $\frac{\pm 3}{5+4 \sin x}$.

If $y = \operatorname{Sin}^{-1}\left(\frac{4+5 \sin x}{5+4 \sin x}\right)$,then find $\frac{dy}{dx}$.

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