If y = sin^{-1} left( frac{5x + 12sqrt{1-x^2} | vedclass

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(A) Given $y = \sin^{-1} \left( \frac{5}{13}x + \frac{12}{13}\sqrt{1-x^2} ight)$. Let $x = \sin 	heta$,then $\sqrt{1-x^2} = \cos 	heta$. Let $\cos

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$\frac{1}{\sqrt{1-x^2}}$

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(A) Given $y = \sin^{-1} \left( \frac{5}{13}x + \frac{12}{13}\sqrt{1-x^2} ight)$. Let $x = \sin 	heta$,then $\sqrt{1-x^2} = \cos 	heta$. Let $\cos \alpha = \frac{5}{13}$,then $\sin \alpha = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$. Substituting these into the equation: $y = \sin^{-1} (\sin 	heta \cos \alpha + \cos 	heta \sin \alpha)$ $y = \sin^{-1} (\sin(	heta + \alpha))$ $y = 	heta + \alpha$ Since $	heta = \sin^{-1} x$ and $\alpha = \cos^{-1} (\frac{5}{13})$ (a constant),$y = \sin^{-1} x + \cos^{-1} (\frac{5}{13})$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx} (\sin^{-1} x) + \frac{d}{dx} (\cos^{-1} (\frac{5}{13}))$ $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} + 0 = \frac{1}{\sqrt{1-x^2}}$.

If $y = \sin^{-1} \left( \frac{5x + 12\sqrt{1-x^2}}{13} \right)$,then $\frac{dy}{dx} = $

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