If sin (x+y)+cos (x+y)=sin left[cos ^{-1}left | vedclass

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(B) Let $\cos ^{-1}\left(\frac{1}{3}ight)=	heta$,then $\cos 	heta=\frac{1}{3}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \s

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(B) Let $\cos ^{-1}\left(\frac{1}{3}ight)=	heta$,then $\cos 	heta=\frac{1}{3}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt{1-\left(\frac{1}{3}ight)^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.Thus,the given equation is $\sin (x+y)+\cos (x+y)=\frac{2\sqrt{2}}{3}$.Since $\frac{2\sqrt{2}}{3} \approx 0.9428$,and the maximum value of $\sin(x+y)+\cos(x+y)$ is $\sqrt{2} \approx 1.414$,this equation represents a constant value for the expression $(x+y)$.Let $x+y = C$,where $C$ is a constant.Differentiating both sides with respect to $x$:$\frac{d}{dx}(x+y) = \frac{d}{dx}(C)$$1 + \frac{dy}{dx} = 0$$\frac{dy}{dx} = -1$.

If $\sin (x+y)+\cos (x+y)=\sin \left[\cos ^{-1}\left(\frac{1}{3}\right)\right]$,then $\frac{d y}{d x}=$

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