If y = cos^{-1}left( frac{3cos x + 4sin x}{5} | vedclass

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(C) Given $y = \cos^{-1}\left( \frac{3}{5}\cos x + \frac{4}{5}\sin x \right)$.Let $\frac{3}{5} = \sin \alpha$ and $\frac{4}{5} = \cos \alpha$,where $\

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(C) Given $y = \cos^{-1}\left( \frac{3}{5}\cos x + \frac{4}{5}\sin x ight)$.Let $\frac{3}{5} = \sin \alpha$ and $\frac{4}{5} = \cos \alpha$,where $\alpha = \sin^{-1}\left( \frac{3}{5} ight)$.Then $y = \cos^{-1}(\sin \alpha \cos x + \cos \alpha \sin x)$.Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $y = \cos^{-1}(\sin(x+\alpha))$.Since $\cos^{-1}(\sin 	heta) = \frac{\pi}{2} - 	heta$,we have $y = \frac{\pi}{2} - (x+\alpha) = \frac{\pi}{2} - x - \alpha$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2} - x - \alpha) = 0 - 1 - 0 = -1$.

If $y = \cos^{-1}\left( \frac{3\cos x + 4\sin x}{5} \right)$,then $\frac{dy}{dx} = $

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